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THE WESTERN CANADA SERIES \^ ^-^ 

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SCHOOL GEOMETRY 



BY 

H. S. HALL, M.A. 

AND 

F. 11. STEVENS, M.A. 



cAuthorized by the 'Departments of Education 
for ^Manitoba, Saskatchewan, oAlberta and 
British Columbia. 



TORONTO 
THE MACMILLAN COMPAN'Y OF CANADA, LIMITED 

19 19 



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Q 4 - a 'y - S. T 



^^iojo^ji^ 



I'opYRiGHT. Canada, 191S, 
By the MACMILLAN COMTANV OF CANADA, LIMITED. 



CONTENTS 



PART I 

t PAGE 

Axioms. Definitions. Postulates ...... 1 

Hypotiiktu'al Con.structions H 

i.n'troductory .......... 8 

Symbols and Abbreviations ....... 9 

Lines and Angles. 

Theorem 1. [Euc. I. 13.] The adjacent angles which one 
straight lino makes with another straight hne on one side of 
it are together equal to two right angles. 10 

Cor. 1. If two straight lines cut one another, the four 
angles so formed are together equal to four right angles. 11 

Cor. 2. When any number of straight lines meet at a 
point, the sum of the consecutive angles so formed is equal to 
four right angles. 11 

Cor. 3. (i) Supplements of the same angle are equal. 
(ii) Complements of the same angle are equal. 1 1 

Theorem 2. [Euc. I. 14.] If, at a point in a straight line, two 
other straight lines, on opposite sides of it, make the adja- 
cent angles together equal to two right angles, then these 
two straight lines are in one and the same straight line. 12 

Theorem 3. [Euc. I. 15.] If two straight lines cut one an- 
other, the vertically opposite angles are equal. 14 

Triangles. 

Definitions Ifl 

The Comparison of Two Triangles 17 

Theorem 4. [Euc. I. 4.] If two triangles have two sides of 
the one equal to two sides of the other, each to each, and the 
angles included by those sides equal, then the triangles are 
equal in all respects. 18 

Theorem .5. (Euc. I. 5.] The angles at tlio base of an isosceles 
triangle are equal. 20 

dm. 1. If tl'.e equil sides of an isosceles triangle are pro- 
duced, the exterior angles at the base are equal. 21 
Cor. 2. If a triangle is equilateral, it is also equiangular. 21 
iii 



^ 



iv CONTENTS 



Theorkm 6. [Euc. I. 6.] If two angles of a triangle arc equal 
to one another, then the sides which are opposite to the equal 
angles are equal to one another. 22 

Theorem 7. [Euc. I. 8.] If two triangles have the three sides 
of the one equal to the three sides of the other, each to each, 
they are equal in all respects. 24 

Theorem 8. [Euc. I. 1(5.] If one side of a triangle is pro- 
duced, then the exterior angle is greater than either of the in- 
terior opposite angles. 28 

Cor. 1. Any two angles of a triangle are together less than 
two right angles. 29 

Cor. 2. Every triangle must have at least two a(^te angles. 29 
Coi{. .3. Only one perpendicular can be drawn to a straight 
hnc from a given point outside it. 29 

Theorem 9. [Euc. I. 18.] If one side of a triangle is greater 
than another, then the angle opposite to the greater side is 
greater than the angle opposite to the less. 30 

Theorem 10. [Euc. I. 19.] If one angle of a triangle is greater 
than another, then the .side opposite to the greater angle is 
greater than the side opposite to the less. 31 

Theorem 11. [Euc. I. 20.] Any two .sides of a triangle arc 
together greater than the third side. 32 

Theorem 12. Of all straight lines from a given point to a given 
straight line the perpendicular is the least. 33 

Cor. 1. If OC is the shortest straight line from O to the 
straight line AB, then OC is perpendicular to AB. 33 

CoR. 2. Two obliques, OP, OQ, which cut AB at e()u il 
distances from C the foot of the perpendicular, are equal. 33 

Cor. 3. Of two obliques OQ, OR, if Ofi cuts .l/i at the 
greater distance from C the foot of the perpendicular, then 
OR is greater than OQ. 33 

Parallels. 

Playfairs Axiom 35 

Theorem 13. [Euc. I. 27 and 28.] If a straight line cuts two 
other straight lines so as to make (i) the alternate angles 
• equal, or (ii) an exterior angle equal to the interior opposite 
angle on the same side of the cutting line, or (iii) the interior 
angles on the same side equal to two richt angles; then in 
each case the two straight lines are parallel. 36 

Theorem 14. [Euc. I. 29.] If a straight line cuts two parallel 
lines, it makes (i) the alternate angles equal to one another; 
(ii) the exterior angle equil to the interior opposite angle on 
the same side of the cuttin<: line ; fiii) the two interior angles 
on the same side together e(iu il to two right angles. 38 



CONTEXTS V 

PAGE 

Parallels Illustrated by Rotation' 39 

Theorem 15. [Euc. I. 30.] Straight lines which are parallel to 

the same straight line are parallel to one another. 40 

Hypothetical Constrtjctiox ....... 40 

Triangles continued. 

Theorem 1G. [Euc. I. 32.] The three angles of a triangle are 
together equal to two right angles. 42 

CoR. 1. All the interior angles of any rectilineal figure, 
together with four right angles, are equal to twice as man}' 
right angles as the figure has sides. 44 

CoR. 2. If the sides of a rectilineal figure, which has no re- 
entrant angle, are produced in order, then all the exterior 
angles so formed are together equal to four right angles. 46 

Theorem 17. (Euc. I. 26.] If two triangles have two angles 
of one equal to two angles of the other, each to each, and any 
side of the first equal to the corresponding side of the other, 
the triangles are equal in all respects. 48 

Ox the Identical Equality of Triangles .... 50 

Theorem 18. Two right-angled triangles which have their 
hypotenuses equal, and one side of one equal to one side of 
the other, are equal in all respects. 51 

Theorem 19. [Euc. I. 24.] If two triangles have two sides of 
the one equal to two sides of the other, each to each, but the 
angle included by the two sides of one greater than the angle 
included by the corresponding sides of the other ; then the 
base of that which has the greater angle is greater than the 
base of the other. 52 

Converse of Theorem 19 . .' . . . .53 

Parallelograms. 

Definitions 56 

Theorem 20. [Euc. I. 33.] The straight fines which join the 
extremities of two equal and parallel straight lines towards 
the same parts are themselves equal and parallel. 57 

Theorem 21. [Euc. I. .34.] The op^ibsite sides and angles of a 
parallelogram are equal to one another, and each diagonal bi- 
sects the parallelogram. .58 

Coft. 1. If one angle of a parallelogram is a right angle, all 
its angles are right angles. 59 

CoR. 2. All the sides of a .square are equal ; and all its 
angles are right angles. 59 

CoR. 3. The diagonals of a parallelogram bisect one an- 
other. 59 



Vi CONTEXTS 



Theorem 22. If there are tliree or more iKirallel straifilit lines, 
and the intereepts made by tlieni on any transversal are equal, 
then the eorresponding intercepts on any other transversal 
are also equal. (52 

Cor. In a triangle ABC, if a set of lines Pp, Qq, Rr, . . ., 
drawn parallel to the base, divide one side .4 B into equal parts, 
they also divide the other side AC into e(jual parts. 03 

Diagonal Scales 66 

Practical Geometry. Problems. 

Inthoductiox. Necessary Instruments .... 69 
Problems on Lines and Angles. 

Problem 1. To bisect a given angle. 70 

Problem 2. To bi.sect a given .'straight line. 71 

Problem 3. To draw a straight line perpendicular to a given 

straight line at a given point in it. 72 

Problem 4. To draw a straight line perpendicular to a given 

straight line from a given external jjoint. 74 

Problem .5. At a given point in a given straight line to make 

an angle equal to a given angle. 7i) 

Problem 6. Through a given point to draw a straight line 

parallel to a given straight line. 77 

Problem 7. To divide a given straight line into any number 

of equal parts. 78 

The Construction of Triangles. 
Problem 8. To draw a triangle, having given the lengths of 

the three sides. SO 

Problem 9. To construct a triangle having given two sides 

and an angle opposite to one of them. 82 

Problem 10. To construct a right-angled triangle having given 

the hypotenuse and one side. 83 

The Construction of Quadril.\terals. 
Problem 11. To construct a quadrilateral, given the lengths 

of the four sides, and one angle. SO 

Problem 12. To construct a jiarallelogram having given two 

adjacent sides and the included angle. 87 

Proble.m 13. To construct a square on a given side. 88 

Loci. 

Problic.m 14. To find the locus of a point P which moves so 
that its distances from two fixed points A and B are always 
equal to one another. 91 



CONTEXTS Vll 



Problem 15. 'Bo find the locus of a point P which moves so 
that its perpendicular distances from two given straight lines 
AB, CD are equil to one another. 92 

Intersectiox of Loci 93 

The Coxcurrence of Str.ught Lines in a Triaxgle. 

L The perpendiculars drawn to the sides of a triangle from 

their middle points are concurrent. 90 

II. I'he bisectors of the angles of a triangle arc concurrent. 97 

11.'/. Tlie bisectors of an interior angle at one verte.x of a triangle 
and of the exterior angles at the other vertices are con- 
current. 97 

III. The medians of a triangle are concurrent. 98 

Cor. The three medians of a triangle cut one another at a 
point of trisection, the greater segment in eich being towards 
the angular point. 98 

IV. The perpendiculars drawn from the vertices of a triangle to 

the opposite sides are concurrent. 99 

PART II 
Areas. 

Definitions 101 

Theorem 23. Area of a Rectaxgle. 102 

Theorem 24. [Euc. I. 35.] Parallelograms on the same base 

and between the same parallels are equal in area. lOG 

Area of a Parallelogram ....... 107 

Theorem 25. Area of a Triangle. 108 

Theorem 26. [Euc. I. 37.] Triangles on the same base and 
between the same parallels (hence, of the same altitude) are 
equal in area. 110 

Theorem 27. [Euc. I. 39.] If two triangles are equal in area, 
and stand on the same base and on the same side of it, they 
are between the same parallels. 110 

Theorem 28. Area of (i) A Trapezium. 114 

(ii) Any Quadrilateral. 114 

Area of any Rectilineal Figure 116 

Theorem 29. [Euc. I. 47. Pythagoras's Theorem.] In a 
right-angled triangle the square described on the hypotenuse 
is equal to the sum of the squares described on the other two 
sides. 120 

Experimental Proofs of Pythagoras's Theorem . . 122 
Theorem 30. [Euc. I. 48.] If the square described on one side 
of a triangle is equal to the sum of the squares described on 
the other two sides, then the angle contained by these two 
sides is a right angle. 124 



Vlll CONTENTS 



Problem 16. To draw squares whose areas shall be respectively 
twice, three times, four times, . . . , that of a given square. 12G 

Geometrical Illustration of Algebraic Identitias . . . 128 9 

Theorem 31. [Euc. II. 12.] In an obtuse-angled triangle, the 
square on the side subtending the obtu.se angle is equ:d to the 
sum of the squares on the .sides containing the obtu.>^e angle 
together with twice the rectangle contained by one of those 
sides and the projection of the other side upon it. 130 

Theorem 32. [Euc. II. 13.] In every triangle the square on 
the side subtending an acute angle is equal to the sum of the 
squares on the sides containing that angle diminished by 
twice the rectangle contained by one of those sides and the 
projection of the other side upon it. 131 

Theore.m 33. In any triangle the sum of the squares on two 
sides is equal to twice the square on half the third side 
together with twice the square on the median which bisects the 
third side. 133 

Problems on Areas. 

Problem 17. To describe a parallelogram equal to a given 

triangle, and having one of its angles equal to a given angle. 135 
Problem 18. To draw a triangle equal in area to a given 

quadrilateral. 137 

Problem 19. To draw a parallelogram equal in area to a given 

rectilineal figure, and having an angle equal to a given angle. 138 
Proble.m 20. To draw a square e(]ual in area to a given 

rectangle. 139 

PART III 

The Circle. Definitions and First Principles. 143 

Symmetry. Symmetrical Properties of Circles . . 14.') 
Properties of Equal Circles • 147 

Chords. 

Theorem 34. [Euc. III. 3.] If a straight line drawn from 
the centre of a circle bi-sects a chord which does not jiass 
through the centre, it cuts the chord at right angles. 

Conversely, if it cuts the chord at right angles, it bi.sects it. 148 
Cor. 1. The straight line which bi.sects a chord at right 
angles passes through the centre. 140 

Cor. 2. A straight line cannot meet a circle at more than 
two points. 1-19 

Cor. 3. A chord of a circle lies wholly within it. 149 



CONTENTS IX 



PAGE 



I 



Theorem 35. One circle, and only one. can p.iss through any 
three points not in the same straight line. 150 

Cou. 1. The size and position of a circle are fully deter- 
mined if it is known to pass through three given points. 151 

Cor. 2. Two circles cannot cut one another in more than 
two points without coinciding entirely. 151 

Hypothetical Construction 151 

Theorem 36. [Euc. III. 9.] If from a point within a circle 
more than two equal straight lines can be drawn to the cir- 
cumference, that point is the centre of the circle. 152 
Theoreai 37. [Euc. III. 14.] Ecjual chords of a circle are equi- 
distant from the centre. 

Conversely, chords which are equidistant from the centre 
are equal. 154 

Theorem 38. [Euc. III. 15.] Of any two chords of a circle, 
that which is nearer to the centre is greater than one more 
remote. 

Conversely, the greater of two chords is nearer to the 
centre than the less. 156 

Cor. The greatest chord in a circle is a diameter. 157 

Angles in a Circle. 

Theorem 39. [Euc. III. 20.] The angle at the centre of a 
circle is double of an angle at the circumference standing on 
the same arc. 158 

Theorem 40. [Euc. III. 21.) Angles in the same segment of a 
circle are equal. 162 

Converse of Theorem 40. Equal angles standing on the 
same base, and on the same side of it, have their vertices on 
an arc of a circle, of which the given base is the chord. 1G3 

Theorem 41. [Euc. III. 22.] The opposite angles of any 
quadrilateral inscribed in a circle are together equal to two 
right angles. 1C4 

Converse OF Theorem 41. If a pair of opposite angles of 
a quadrilateral are supplementary, its vertices are concyclic. 165 

Tangency. 

Definitions ant) First Principles 168 

Theorem 42. The tangent at any point of a circle is perpen- 
dicular to the radius drawn to the point of contact. 170 

Cor. 1. One and only one tangent can be drawn to a 
circle at a given point on the circumference. 170 



X CONTENTS 

PAOB 

Cor. 2. Tho pcrpcndicul ir to :i taimcnt at its point of 
contact passes through the cenlro. 170 

Cor. 3. The radius drawn perpendicular to the tangent 
passes through the point of contact. 170 

Theorem 43. Two tangents can be drawn to a circle from an 
external point. 171 

Cor. The two tangents to a circle from an external ])oint 
are equal, and subtend equal angles at the centre. 171 

TiiKOREM 44. If two circles touch one another, t!ie centres and 
the point of contact are in one straight line. 173 

CoR. 1. If two circles touch externally, the distance be- 
tween their centres is equal to the sum of their radii. 173 

CoR. 2. If two circles touch internally, the distance be- 
tween their centres is equal to the difference of their radii. 173 
Theorem 45. [Euc. III. 32.] The angles made by a tangent 
to a circle with a chord drawn from the point of contact are 
respectively equal to the angles in the alterna(e segments of 
the circle. 175 

Problems. 

Geometrical Ax.\lys!.s ........ 177 

Problem 21. Given a circle, or an arc of a circle, to find its 

centre. 178 

Problem 22. To bi.sect a given arc. 178 

Problem 23. To draw a tangent to a circle from a given ex- 
ternal point. 179 
Problem 24. To draw a common tangent to two circles. 180 

The Construction of Circles 183 

Problem 25. On a given straight line to describe a segment of 
a circle which shall contain an angle equal to a given angle. 185 

Cor. To cut off from a given circle a segment containing 
a given angle, it is enough to draw a tangent to the circle, 
and from the point of contact to draw a chord making with 
the tangent an angle equal to the given angle. 18G 

Circles in Relation to Rectilineal Figures. 

Definitions .......... 187 

PuouLE.M 2G. To circumscribe a circle about a given triangle. 188 

Problem 27. To in.scribe a circle in a given triangle. ISO 

Problem 28. To draw an escrilx-d circle of a given triangle. 190 
Puoble.m 29. In a given circle to in.scribe a triangle ('(lui- 

angular to a given triangle. 191 



CONTENTS XI 



Problem 30. About a given rirclc to circumscribe a triangle 

equiangular to a given triangle. 192 

Problem 31. To draw a regular polygon (i) in (ii) about a 

given circle. 195 

Problem 32. To draw a circle (i) in (ii) about a regular polygon. 19G 

Circumference and Area of a Circle 197 



PART IV 
Proportion. 

Definitioxs .^xd First Prixciple.s 203 

Introductory Theorems I.-\T 205 

Proportional Division of Straigiit Lines. 

Theorem 46. [Euc. VI. 2.] A straight line drawn parallel to 
one side of a triangle cuts the other two sides, or those sides 
produced proportionally. 210 

Theorem 47. [Euc. \l. 3 and A.] If the vertical angle of a 
triangle is bi.sected internally or externally, the bisector 
divides the base internally or externally into segments which 
have the same ratio as the other sides of the triangle. 

Conver.sely, if the base is divided internally or externally 
into segments proportional to the other sides of the tiiangle, 
the line joining the point of section to the vertex bisects the 
vertical angle internally or externalh'. 212 

Proportional Areas. 

Theorem 48. [Euc. VI. 1.] The areas of triangles of equal 
altitude are to one another as their bases. 216 

Cor. The areas of parallelograms of equal altitude are to 
one another as their bases. 217 

Proportional Arcs and Angles. 

Theorem 49. [Euc. VI. 33.] In equal circles, angles, whether 
at the centres or circumferences^ have the same ratio as the 
arcs on which they stand. " 218 

Cor. In equal circles, sectors have the same ratio as 
their angles. 218 

Similar Figures. Definitions 219 

Similar Triangles. 

Theore.m 50. [Euc. VI. 4.] If two triangles are equiangular 
to one another, their corresponding sides are i)roportional. 220 



Xii CONTENTS 



Theorem 51. [Euc. VI. "i.] If two triangles have their sides 

f)roportional when tukeu iii order, the triangles are equiangu- 
ar to one another, and those angles are equal which are 
opposite to corresponding sides. 221 

Theorem 52. [Euc. VI. 6.] If two triangles have one angle 
of the one equal to one angle of the other, and the sides about 
the equal angles proportionals, the triangles are similar. 224 

Theorem 53. [Euc. VI. 7.) If two triangles have one angle of 
the one equal to one angle of the other, and the sides about 
another angle in one proportional to the corresijonding sides of 
the other, then the third angles are either equal or supple- 
mentary; and in the former case the triangles are similar. 225 

Theorem 54. [Euc. VI. 8.] In a right-angled triangle, if a 
perpendicular is drawn from the right angle to the hypote- 
nuse, the triangles on each side of it are similar to the whole 
triangle and to one another. 227 

Theorem 55. [Euc. VI. 19.] The areas of similar triangles are 
proportional to the squares on corresponding sides. 229 

Theorem 56. [Euc. III. 35 and 36.] If any two chords of a 
circle cut one another internally or externally, the rectangle 
contained by the segments of one is equal to the rectangle 
contained by the segments of the other. 231 

Cor. If from an external point a secant and n tangent are 
drawn to a circle, the rectangle contained by the whole se- 
cant and the part of it outside the circle is equal to the 
square on the tangent. 232 

Problems. 

Problem 33. To find the fourth proportional to throe given 
straight lines. 235 

Problem 34. To find the third proportional to two given 
straight lines. 2.''5 

Problem 35. To divide a given straight lino internally and 
externally in a given ratio. 23(5 

Pii( )HLi:.M 36. To find the mean proportional between two given 
straight linos. 237 

Similar Polygons. 

Theorem 57. Similar jiolygons can bo divided into the .same 
number of similar triangles ; and the lines joining corre.spond- 
ing vertices in each figure are proportional. 240 

Problem 37. On :i side of given longlli to draw a figure similar 
to a given rectilineal liguro. 242 



CONTENTS xiii 



Theorem 58. Any two similar rectilineal figures may be so 
placed that the lines joining corresponding vertices are con- 
current. 243 

Theorem 59. [Euc. VI. 20.] The areas of similar polygons 
are proportional to the squares on corresponding sides. 24(3 

Theorem 60. [Euc. VI. 31.] In a right-angled triangle, any 
rectilineal figure described on the hypotenuse is equal to the 
sum of the two similar and similarly described figures on the 
sides containing the right angle. 249 

Problem 38. To draw a figure similar to a given rectilineal 
figure, and equal to a given fraction of it in area. 251 

Miscellaneous Theorems. 

Theorem 61. If the vertical angle of a triangle is bisected by 
a straight line which cuts the ba.sc, the rectangle contained by 
the sides of the triangle is equal to the rectangle contained by 
the segments of the ba.se, together with the .square on the 
straight line which bisects the angle. 253 

Theorem 62. If from the vertical angle of a triitngle a straight 
line is drawn perpendicular to the base, the rectangle con- 
tained by the sides of the triangle is equal to the rectangle 
contained by the perpendicular and the diameter of the cir- 
cum-circle. 254 

Theorem 63. [Ptolemy's Theorem.] The rectangle contained 
bj^ the diagonals of a quadrilateral inscribed in a circle is 
equal to the sum of the two rectangles contained by its 
opposite sides. 255 

Examples, Parts I-IV . 257 

Answers to Numerical Exercises 261 



GEOMETRY 



PART I 

AXIOMS 

All mathematical rca^^oiiing is founded on certain simple 
principles, the truth of which is so evident that they are 
accepted without proof. These self-evident truths are called 
Axioms. 

For instance : 

Things which arc equal to the same thing arc equal to one 
another. 

The following axioms, corresponding to the first four Rules 
of Arithmetic, are among those most commonly used in 
geometrical reasoning. 

Addition. // equals are added to equals, the sums are equal. 

Subtraction. If equals are taken from equals, the remainders 
are equal. 

Multiplication. Things which ore the same midtiples of 
equals are equal to one another. 

For instance: Doubles of equals arc equal to one another. 

Division. Things which are the same parts of equals are 
equal to one another. 

For instance : Halves of equals are equal to one another. 

The above Axioms are given as instances, and not as a 
complete list, of those which will be used. They are said to 
B 1 



2 GEOMETRY 

be general, because they applj^ equally to magnitudes of all 
kinds. Certain special axioms relating to geometrical magni- 
tudes only will be stated from time to time as they are 
required. 

Definitions and First Principles 

Every beginner knows in a general way what is meant by a 
point, a line, and a surface. But in geometry these terms 
are used in a strict sense which needs some explanation. 

1.. A point has position, but is said to have no magnitude. 

This means that we are to attach to a point no idea of size either 
as to length or breadth, but to think only where it is situated. A dat 
made wth a sharp pencil may be taken as roughly representing a 
point ; but small as such a dot maj' be, it still has some length and 
breadth, and is therefore not actually a geometrical point. The 
smaller the dot however, the more nearly it represents a point. 

2. A line has length, but is said to have no breadth. 

A line is traced out by a moving point. If the point of a pencil is 
moved over a sheet of paper, the trace left represents a line. But 
such a trg,ce, however finely drawn, has some degree of breadth, and 
is therefore not itself a true geometrical line. The finer the trace 
h^ft by the moving pencil-point, the more nearly will it represent a 
line. 

3. Proceeding in a similar manner from the idea of a line 
to the idea of a surface, we say that 

A surface has length and brcadtli, but no thickness. 
And finally, 

A solid has length, breadth, and thickness. 
Solids, surfaces, lines, and points are thus related to one another: 
(i) A solid is bounded by surfaces. 

(ii) A surface is bounded by lines ; and surfaces meet in lines, 
(iii) A line is bounded (or terminated) by points ; and lines meet 
in points. 



DEFINITIONS 3 

4. A line may be straight or curved. 

A straight line has the same direction from point to point 
throughout its whole length. 

A curved line changes its direction continually from point 
to point. 

Axiom. There can he only one straight line joining two 
given points: that is, 

Two straight lines cannot enclose a space. 

5. A plane is a flat surface, the test of flatness being that 
if an}^ two points are taken in the surface, the straight line 
between them lies wholly in that surface. 

G. When two straight lines meet at a 
point, they are said to form an angle. 

The straight lines are called the arms of the 
angle ; the point at which they meet is its vertex. 

The magnitude of the angle may be thus 
explained : 

Suppose that the arm OA is fixed, and that OB turns about 
the point (as shewn by the arrow). Suppose also that OB 
began its turning from the position OA . Then the size of the 
angle AOB is measured by the amount of turning required to 
bring the revolving arm from its first position OA into its 
subsequent position OB. 

Observe that the size of an angle docs not in anj^ waA depend on 
the length of its arms. _^ 

Angles which lie on either side 
of a common arm ai-e said to be 
adjacent. 

For example, the angles AOB, BOC, 
which have the common arm OB, are 
adjacent. 





GEOMETRY 



When two straight Hnes such asAB, CD 
cross one another at 0, the angles CO A, _ 
BOD are said to be vertically opposite. ^ 
The angles AOD, COB are also vertically 
opposite to one another. 

7. When one straight line stands on an- 
other so as to make the adjacent angles equal 
to one another, each of the angles is called a 
right angle ; and each line is said to be per- 
pendicular to the other. 



"O A 



O A 



Axioms, (i) // is a point in a straight line AB, then a line 
OC, which turns about Ofrom the position OA to the position OB, 
must pass through one position, and only one, in which it is 
perpendicular to AB. 

(ii) All right angles are equal. 

A right angle is divided into 00 equal parts called degrees (°) ; 
each degree into 60 equal parts called minutes (') ; each minute into 
60 equal parts called seconds (")• 

In the above figure, if OC revolves about from the 
position OA into the position OB, it turns through two right 
angles, or 180°. 

If OC makes a complete revolution about 0, starting from 
OA and returning to its original position, it turns through 
four right angles, or 360°. 8 

8. An angle which is less than one right 
angle is said to be acute. 

That is, an acute angle is less than 00". 

9. An angle which is greater than 
one right angle, but less than two 
right angles, is said to be obtuse. 

That is, an obtuse angle lies be- 
tween 90° and 180°. 




DEFINITIONS 

10. If one arm OB of an angle 
turns until it makes a straight line with 
the other arm OA, the angle so formed 
is called a straight angle. 

A straight angle = 2 right angles = 180"" 

11. An angle which is greater 
than iico right angle?, but less than 
four right angles, is said to be 
reflex. B 

That is, a reflex angle lies between 180'' and 360°. 

Note. When two straight lines meet, two angles are formed, one 
greater, and one less than two right angles. The first arises by 
Supposing OB to have revolved from the position OA the longer way 
round, marked (i) ; the other by supposing OB to have revolved the 
shorter way round, marked (ii). Unless the contrarj' is stated, the 
angle between two straight linos will be considered to be that which 
is less than two right angles. 

12. Any portion of a plane surface bounded bj' one or 
more lines is called a plane figure. 

13. A circle is a plane figure contained 
by a line traced out by a point which 
moves so that its distance from a certain 
fixed point is alwaj^s the same. 

Here the point P moves so that its distance 
from the fixed point is always the same. 

The fixed point is called the centre, and the bounding line 
is called the circumference. "^ 

14. A radius of a circle is a straight line drawn from the 
centre to the circumference. It follows that all radii of a 
circle are equal. 

15. A diameter of a circle is a straight line drawn through 
the centre, and terminated both ways by the circumference. 





6 GEOMETRY 

10. An arc of a circle is anj^ part of the circumference. 

17. A semi-circle is the figure bounded 
l)y a diameter of a circle and the part of 
the circumference cut off by the diameter. 

18. To bisect means to divide into two equal parts. 

Axioms, (i) // a point O moves -i- 

froin A to B along the straight line AB, 

it must pass through o}ie position in which it divides A B into 

two equal parts. 

That is to say : 

Every finite straight line has a point of bisection. 

(ii) If a line OP, revolving about O, turns 
from OA to OB, it must pass through one 
position in which it divides the angle AOB 
into two equal parts. ^ . 

That is to say : 

Every angle may be supposed to have a line of bi.'ieeti<m. 

Hypothetical Constructions 

From the Axioms attached to Definitions 7 and 18, it 
follows that we may suppose 

(i) A straight line to be draivn perpoulieular to a given 
straight Hue from any point in it. 

(ii) A finite straight line io be bisected at a point. 
(iii) An angle to be bisected by a line. 

Superposition and Equality 

Axiom. Magnitudes which can be made to coincide with one 
another are equal. 

This axiom implies tliat any line, anplr. or fipfurf^ may be takrn 
up from its position, and without chango in size or form, hiid down 



i 



i 



POSTULATES 7 

upon a second line, angle, or figure, for the purpose of comparison, 
and it states that two such magnitudes are equal when one can lie 
exactly placed over the other without overlapping. 

This process is called superposition, and the first magnitude is said 
to be applied to the other. 

Postulates 

In' order to draw geometrical figures certain instruments 
are required. These are, for the purposes of this book, (i) a 
straight ruler, (ii) a pair of compasses. The following Postu- 
lates (or requests) claim the use of these instruments, and 
assume that with their help the processes mentioned below 
may be duly performed. 

Let it be granted : 

1. That a straight line may be drawn from any one point to 
any other point. 

2. That a finite {or terminated) straight line may he 
PRODUCED {that is, prolonged) to any length in that straigJit 
line. 

3. That a circle may he drawn with any point as centre and 
with a radius of any length. 

Notes, (i) Postulate 3, as stated above, im- 
plies that we may adjust the compasses to the 
length of any straight line PQ, and ^vith a ra- 
dius of this length draw a circle with any point 
O as centre. That is to say, the compasses 
may be used to Iransfer distances from one part 
of a diagram to another. ^ 

(ii) Hence from AB, the greater of two 
straight lines, we may cut off a part equal to 
PQ the less. 

For if with centre A, and radius equal a 
to PQ, we draw an arc of a circle cutting 
AB Sit X, it is obvious that AZ is equal ~_ 
to PQ. P 




Q 



GEOMETRY 



Introductory 



1. Plane geometry deals with the properties of such lines 
and figures as may be drawn on a plane surface. 

2. The subject is divided into a number of separate dis- 
cussions, called propositions. 

Propositions are of two kinds, Theorems and Problems. 

A Theorem proposes to prove the truth of some geometrical 

statement. 

A Problem proposes to perform some geometrical construc- 
tion, such as to draw some particular line, or to construct 
some required figure. 

3. A Proposition consists of the following parts : 

The General Enunciation, the Particular Enunciation, tlie 
Construction, and the Proof. 

(i) The General Enunciation is a preliminary statement, 
describing in general terms the purpose of the proposition. 

(ii) The Particular Enunciation rei)eats in special terms 
the statement already made, and refers it to a diagram, which 
enables the reader to. follow the reasoning moi-e easilj'. 

(iii) The Construction then directs the drawing of such 
straight lines and circles as may lie required to efTect the 
purpose of a problem, or to prove the truth of a theorem. 

(iv) The Proof shews that the object proposed in a prob- 
lem has been accomplished, or that the property statetl in a 
tlieorem is true. 

4. The letters q.e.d. are appended to a theorem, and stand 
iuv Quod erat Demonstrandum, which ivas to he jrrorrd. 



INTRODUCTORY 9 

5. A Corollary is a statement the truth of which follows 
readily from an established proposition ; it is therefore 
appended to the proposition as an inference or deduction, 
which usually requires no further proof. 

6. The following symbols and abbreviations are used in 
the text of this book : 



In Part I. 






.'. for therefore, 


Z 


for angle, 


= " is, or are, equal to, 


A 


" triangle. 


After Part I. 






pt. for point, 


perp. 


for perpendicular. 


St. line " straight line, 


par°> 


" parallelogram. 


rt. Z " right angle. 


rectil, 


, " rectilineal. 


par* (or |I) " parallel. 


O 


" circle. 


sq. " square, 


Qce 


" circumference ; 



and all obvious contractions of commonly occurring words, 
such as opp., adj., diag., etc., for opposite, adjacent, diagonal, 
etc. 

[For convenience of oral work, and to prevent the rather common 
abuse of contractions by beginners, the above code of signs has been 
introduced gradually, and at first somewhat sparingly.] 

In numerical examples the following abbreviations will 
be used, 
m. for metre, cm. for centimetre, 

mm. " millimetre. km. " kilometre. 

Also inches are denoted by the sj^mbol (")• 
Thus 5" means 5 inches. 



10 



GEOMETRY 



ON LINES AND ANGLES 

Theorem 1. [Euclid I. L3] 

The adjacent angles ivhich one straight line makes with an- 
other straight line on one side of it, are together equal to two 
right angles. 




B 

Let the straight Hne CO make with the straight Hne AB 
the adjacent A AOC, COB. 

It is required to prove that the A AOC, COB are together equal 
to two right angles. 

Suppose OD is at right angles to BA. 
Proof. Then the A AOC, COB together 

= the three A AOC, COD, DOB. 
Also the A AOD, DOB together 

= the three A AOC, COD, DOB. 
:. the A AOC, COB together = the A AOD, DOB 

= two right angles. 

Q.E.D. 

PROOF BY ROTATION 

Suppose a straight line revolving about O turns from the position 
OA into the position OC, and thence into tlie position Oli; that is, 
let the revolving line turn in succession through the A A(^C, COB. 

Now in passing from its first position OA to its final position 
Oli, the revolving line turns through two right angles, for .'\()li is a 
straight line. 

Hence the A AOC, COli togcUier -^ two right niiglts. 



LINES AND ANGLES 



11 



Corollary 1. If two straight lines 
cut one another, the four angles so formed 
are together equal to four right angles. A 

For example, 
Z BOD + Z DOA + Z AOC + Z COB = 4 righl angles. 




Corollary 2. When any number of 
straight lines meet at a point, the sum of 
the consecutive angles so formed is equal 
to four right angles. 




For a straight line revolving about O, and turning in succession 
through the A AOB, BOC, COD, DOE, EOA, will have made one 
complete revolution, and therefore turned through four right angles. 



DEFINITIONS 

(i) Two angles whose sum is two right angles are said to 
be supplementary ; and each is called the supplement of the 
other. 

Thus in the Fig. of Theor. 1 the angles AOC, COB are supple- 
mentary. Again the angle 123° is the supplement of the angle 57°. 

(ii) Two angles whose sum is oyie right angle are said to 
be complementary ; and each is^alled the complement of the 
other. 

Thus in the Fig. of Theor. 1 the angle DOC is the complement of 
the angle AOC. Again angles of 34° and 56° are complementary. 

Corollary 3. (i) Supplements of the same angle are equal. 
(ii) Complements of the same angle are equal. 



12 GEOMETRY 



Theorem 2. [Euclid I. 14] 

//, at a point in a straight line, two other straight lines, on 
opposite sides of it, make the adjacent angles together equal to two 
right angles, then these two straight lines are in one and the same 
straight line. 



At in the straight hnc CO let the two straight lines OA, 
OB, on opposite sides of CO, make the adjacent A AOC, 
COB together equal to two right angles : (that is, let the 
adjacent A AOC, COB be supplementary). 

It is required to prove that OB and OA are in the same straight 
line. 

Produce A beyond to any point X : it will be shewn 
that OA' and OB are the same line. 

Proof. Since by construction AOX is a straight line, 

.'. the Z COA is the supplement of the A CO A. Theor.l 
But, by hypothesis, 

the A COB is the supplement of the A COA. 
.: the A COX = the A COB ; 
.". OX and OB are the same line. 

But, by construction, OX is in the same straight line with 
OA;. 

hence OB is also in the same straight line with OA . 

Q.E.D. 



LINES AND ANGLES 13 



EXERCISES 

1. Write down the supplements of onc-halj of a right angle, Jour- 
thirds of a right angle ; also of 46°, 149°, 83°, 101° 15'. 

2. Write down the complpmcnt of two-fiflhs of a right angle ; 
also of 27°, 38° 16', and 41° 29' 30". 

3. If two straight lines intersect forming four angles of which 
one is known to be a right angle, prove that the other tliree are also 
right angles. 

4. In the triangle ABC the angles ABC, ACB are given equal. 
If the side BC is produced both ways, shew that the exterior angles 
so formed are equal. 

5. In the triangle ABC the angles ABC, ACB are given equal. 
If A B and AC are produced beyond the base, shew that the e.xterior 
aiugles so formed are equal. 

Definition. The lines which bisect an angle and the 
adjacent angle made by producing one of its arms are called 
the internal and external bisectors of the given angle. 

Thus in the diagram, OX and OY are 
the internal and external bisectors of the 
angle AOB. 

C 

6. Prove that the bisectors of the adjacent angles which one 
straight line makes with another contain a right angle. That is to 
say, the internal and external hisector&of an angle are at right angles 
to one another. 

7. Shew that the angles AOX and COY in the above diagram 
are complementary. 

8. Shew that the angles BOX and COX are supplementary; 
and also that the angles AOY and BOY are supplementary. 

9. If the angle AOB is 35°, find the angle COY. 




14 GEOMETRY 

Theorem 3. [Euclid I. 15] 

// two straight lines cut one another, the vertically opposite 
angles are equal. 




Let the straight hnes AB, CD cut one another at tlie 
point 0. 
It is required to prove that 

(i) the Z AOC = the Z DOB ; 
(ii) the Z COB = the Z AOD. 

Proof. Because AO meets the straight hne CD, 
.'. the adjacent A AOC, AOD to^vthcr = two right angles ; 
that is, the Z AOC is the supplement of the Z AOD. 
Again, because DO meets the straight line A B, 
.'. the adjacent A DOB, AOD together = two right angles; 
that is, the Z DOB is the supplement of the Z AOD. 
Thus each of the A AOC, DOB is the supplement of the 
Z AOD, 

.: the Z AOC = the Z DOB. 
Similarly, the Z COB = the Z AOD. 

Q.E.l). 

PROOF BY ROTATION 

Suppose the line COD to revolve about O until OC turns into the 
position ()A. Then at the same moment 01) must reach the posi- 
tion OB (for AOB and COD are slrnighl). 

Thus the same amount of turning is required to elose the Z AOC 
as to close the Z DOH. 

.: the Z AOC - the Z /;(7/i. 



LINES AND ANGLES 15 

EXERCISES ON ANGLES 
( Numerical) 

1. Through what angles does the minute-hand of a clock turn in 
(i) 5 minutes, (ii) 21 minutes, (iii) 43^ minutes, (iv) 14 min. 10 sec? 
And how long will it take to turn through (v) 66", (vi) 222°? 

2. A clock is started at noon : through what angles will the hour- 
hand have turned by (i) 3.45, (ii) 10 minutes past 5? And what 
will be the time when it has turned through 1722°? 

3. The earth makes a complete revolution about its axis in 24 
hours. Through what angle will it turn in 3 hrs. 20 min.? 

4. In the diagram of Theorem 3 

(i) If the Z. AOC = 35°, write down (without measurement) the 
value of each of the A COB, BOD, DO A. 

(ii) If the A COB, AOD together make up 250°, find each of the 
A CO A, BOD. 

(iii) If the A AOC, COB, BOD together make up 274°, find each 
of the four angles at O. 

{Theoretical) 

' 5. If from O, a point in A B, two straight lines OC, OD are drawn 
on opposite sides of AB so as to make the angle COB equal to the 
angle AOD ; shew that OC and OD are in the same straight line. 

6. Two straight lines AB, CD cross at O. If OX is the bisector 
of the angle BOD, prove that XO produced bisects the angle AOC. 

7. Two straight lines AB, CD cross at 0. If the angle BOD is 
bisected by OX, and AOC by OY, prove that OX, OY are in the 
same straight line. 

8. If OX bisects an angle AOB, shew that, by folding the dia- 
gram about the bisector, OA may be made to coincide with OB. 

How would OA fall with regard to OB, if 

(i) the Z AOX were greater than the Z XOB ; 
(ii) the Z .40X v/ere less^an the Z XOB? 

9. AB and CD are straight lines intersecting at right angles at 
O ; shew by folding the figure about .4 B, that OC may be made to 
fall along OD. 

10. A straight line ^4 OB is drawn on paper, which is then folded 
about 0, so as to make OA fall along OB ; shew that the crease left 
in the paper is perpendicular to AB^ 



16 



GEOMETRY 



ON TRIANGLES 

1. Any portion of a plane surface bounded by one or 
more lines is called a plane figure. 

The sum of the bounding Unes is called the perimeter of the figure. 
The amount of surface enclosed by the perimeter is called the area. 

2. Rectilineal figures are those which are bounded by 
straight lines. 

3. A triangle is a plane figure bounded by three straight 
lines. 

4. A quadrilateral is a plane figure bounded l)y four 
straight lines. 

5. A polygon is a plane figure bounded by 
more than four straight lines. 

0. A rectilineal figure is said to be 
equilateral, when all its sides are equal; 
equiangular, when all its angles are equal; 
regular, when it is both equilateral and equiangular. 

7. Triangles are thus classified with regard to their sides : 
A triangle is said to be 

equilateral, when all its sides are equal ; 
isosceles, when two of its sides are equal; 
scalene, when its sides are all unequal. 







Equilateral Triangle 



Isosceles Triangle 



Scalene Triangle 



In a triangle .1 BC, tlic letters .1, B, f often ilf- 
note l\w magnitude of the several angles (as meas- 
ured in degrees) ; and the letters n, b, c the h-ngllm 
of the opposite sides (as measured in inches, centi- 
metres, or some other unit of length). 




TRIANGLES 17 

Any one of the angular points of a triangle may be regarded as 
its vertex ; and the opposite side is then called the base. 

In an isosceles triangle the term I'ertex is usually applied to the 
point at which the equal sides intersect ; and the vertical angle is 
the angle included by them. 

8. Triangles are thus classified with regard to their angles : 
A triangle is said to be 

right-angled, when one of its angles is a right angle; 
obtuse-angled, when one of its angles is obtuse; 
acute-angled, when all three of its angles are acute. 
[It will be seen hereafter (Theorem 8. Cor. 1) that every triangle 
must have at least two acute angles. 




Right-angled Triangle Obtuse-angled Triangle Acute-angled Triangle 

In a right-angled triangle the side opposite to the right angle is 
called the hypotenuse. 

9. In any triangle the straight line joining a vertex to the 
middle point of the opposite side is called a median. 

THE COMPARISON OF TWO TRIANGLES 

(i) The three sides and three angles of a triangle are called 
its six parts. A triangle may also be considered with regard 
to its area. 

(ii) Two triangles are said to be equal in all respects, 
when one may be so placed upon the other as to exactly 
coincide with it ; in which case each part of the first triangle 
is equal to the corresponding part (namely that with which 
it coincides) of the other ; and the triangles are equal in area. 

In two such triangles corresponding sides are opposite to 
equal angles, and corresponding angles are opposite to equal sides. 

Triangles which may thus be made to coincide by super- 
position are said to be identically equal or congruent. 



18 GEOMETRY 

Theorem 4. [Euclid I. 4] 

// two triangles have two sides of the one equal to two sides of 
the other, each to each, and the angles included by those sides 
equal, then the triatigles are equal in all respects. 

D 




Let ABC, DEF be two triangles in which 

AB = DE, 
AC = DF, 
and the included angle BAC = the included angle ED F. 
It is required to prove that the A ABC = the A DEF in all 
respects. 

Proof. Apply the A AfiC to the A DEF, 

so that the point A falls on the point D, 
and the side AB along the side DE. 
Then because AB = DE, 
.'. the point B must coincide with the point E. 
And because AB falls along DE, 
and the Z BAC = Z EDF, 
.: AC nnist fall along DF. 
And because AC = DF, 
.'. the point C must coincide with the point F. 

Then since B coincides with E, and C with F, 
.'. the !-ide BC must coincide with the side EF. 
Hence the A ABC coincides with the A DEF, 
and is therefore equal to it in all respects. 

y.E.D. 



CONGRUENT TRIANGLES 19 

Obs. In this Theorem we must carefully observe what is 
given and what is proved. 

[ AB - DE, 

Given that j AC = DF, 

[and the Z BAC = the Z EDF. 
From these data we prove that the triangles coincide on 
superposition. 

[ BC = EF, 

Hence we conclude that the /. ABC = the Z DEF, 

[and the Z ACjB = the Z D/^i?; 
also that the triangles are equal in area. 

Notice that the angles which are proved equal in the two 
triangles are opposite to sides which were given equal. 

A P 

iJoTE. The adjoining: diagram shows 
that in order to make two congruent 
triangles coincide, it may be necessar.y 
to reverse, that is, turn over one of them 
before superposition. 

EXERCISES 

1. Shew that the bisector of the vertical angle of an isosceles triangle 
(i) bisects the base, (ii) is perpendicular to the base. 

2. Let be the middle point of a straight line AB, and let OC be 
perpendicular to it. If P is any point in OC, prove that PA = PB. 

3. Assuming that the four sides of a square A BCD are equal, 
and that its angles are all right a^igles, prove the diagonals AC, 

^iBD equal. 

4. A BCD is a square, and L, M, and A" are the middle points of 
AB, BC, and CD: using a separate figure in each case, prove that 
(i) LM^MN. (ii) AM = DM. (iii) AN = AM. (iv) BN = DM. 

5. ABC is an isosceles triangle: from the equal sides AB, AC 
two equal parts .4 A', AY are cut off, and BY and CX are joined. 
Prove that BY = CX. 




20 GEOMETRY 

Theorem 5. [Euclid I. 5] 
The angles at the base of an isosceles triangle are equal. 

A 



B D C 

Let ABC be an isosceles triangle, in which the side AB = 
the side AC. 

It is required to prove that the Z ABC = the Z ACB. 

Suppose that AD is the line which bisects the Z BAC, and 
let it meet BC in D. 

1st Proof. Then in the A BAD, CAD, 
f BA = CA, 

because! AD is common to both triangles, 

[ and the included /. BAD = the included l CAD; 
:. the triangles are equal in all respects ; Theor. 4. 
so that the Z ABD = the Z ACD. 

Q.E.D. 

2nd Proof. Suppose the A ABC to be folded about AD. 
Then since the Z BAD = the Z CAD, 
.'. AB must fall along AC. 

And since AB = AC, 
.'. B must fall on C, and consequently DB on DC. 
.'. the Z ABD will coincide with the Z ACD, and is therefore 
equal to it. q.e.d. 



^ 




ISOSCELES TRIANGLES 21 



Corollary 1. // the equal sides AB, AC 
of an isosceles triangle are produced, the exterior 
angles EBC, FCB are equal; for they are the 
supplements of the equal angles at the base. 

E' \F 

Corollary 2. // a triangle is equilateral, it is also equi- 
angular. 

Definition. A figure is said to be symmetrical about a 
line wlien, on being folded about that line, the parts of the 
figure on each side of it can be brought into coincidence. 

The straight line is called an axis of symmetry. 

That this may be possible, it is clear that the two parts of the 
figure must have the same size and shape, and must be similarly 
placed with regard to the axis. 

EXERCISES 

/ 1. A BCD is a four-sided figure whose sides are all equal, and the 
diagonal BD is drawn: shew that 

(i) the angle ABD = the angle ADB; 

(ii) the angle CBD = the angle CDB; 

(iiij the angle ABC = the angle ADC. 

\/ 2. ABC, DBC are two isosceles triangles drawn on the same 
base BC, but on opposite sides of it : prove (by means of Theorem 5) 
that 

the angle ABD = the angle ACD. 

H 3. ABC, DBC are two isosceles triangles drawn on the same 
base BC and on the same side of it : employ Theorem 5 to prove that 
the angle ABD = the angle ACD. 

I / 4. AB, AC are the equal sides of an isosceles triangle ABC ; and 
L, M, N are the middle points of AB, BC, and CA respectively: 
prove that (i) LM = Ni\f. (ii) BN = CL. 

(iii) the angle ALM = the angle AXM. 



22 GEOMETRY 

Theorem G. [Euclid I. G] 

// two angles of a tnangle are equal to one another, then the 
sides which are opposite to the equal angles are equal to one 
another. 




Let ABC be a triangle in which 

the Z A5C = the Z ACB. 
It is required to prove that the side AC = the side AB. 
If AC and AB are not equal, suppose that AB is the greater. 
From BA cut off BD equal to AC. 
Join DC. 

Proof. Then in the A DBC, ACB, 

f DB = AC, 

because! BC is common to both, 

[and the included Z DBC = the included Z ACB ; 
.'. the A DBC = the A ACB in area, Theor. 4. 
the part equal to the whole ; which is absurd. 
.•. AB is not unequal to AC ; 
that ifi,AB = AC. 

Q.E.D. 

Corollary. An equiangular triangle is also equilateral. 

In Theorem G we employ an indired method of proof frequently 
used in geometry. It consists in shewing that the theorem cannol he 
untrue; since, if it were, we .should l)e led to some impossible conclu- 
sion. This form of proof is known as Reductio ad Absurdum. 




A THEOREM AND ITS CONVERSE 23 

NOTE ON THEOREMS 5 AND 6 

Theorems 5 and 6 may be verified ex- ^ 

perimentally by cutting out the given / \ 
A ABC, and, after turning it over, fitting ' '< 
it thus reversed into the vacant space left / \ 

in the paper. c. 1, 

B C C B 

Suppose A'B'C to be the original position of the A ABC, and 
let ACB represent the triangle when reversed. 

In Theorem 5, it will be found on applying A to A' that C may be 
made to fall on B', and B on C 

In Theorem 6, on applying C to B' and B to C" we find that A 
will fall on A'. 

In either case the given triangle rercrscd will coincide ^\^th its 
own " trace," so that the side and angle on the left are respectively 
equal to the side and angle on the right. 

NOTE ON A THEOREM AND ITS CONVERSE 

The enunciation of a theorem consists of two clauses. The first 
clause tells us what we are to assume, and is called the hypothesis; 
the second tells us what it is required to prove, and is called the 
conclusion. 

For example, the enunciation of Theorem 5 assumes that in a cer- 
tain triangle ABC the side AB = the side AC: this is the hypothesis. 
From this it is required to prove that the angle ABC = the angle 
ACB: this is the conclusion. 

If we interchange the hypothesis and conclusion of a theorem, 

we enunciate a new theorem which is called the converse of the first. 

For example, in Theorem 5 

it is assumed that AB = AC ; 1 

it is required to prove that the angle ABC = the angle ACB. j 

Now in Theorem 6 

it is assumed that the angle ABC = the angle ACB; ) 
it is required to prove that AB = AC. J 

Thus we see that Theorem 6 is the converse of Theorem o ; for 
the hypothesis of each is the conclusion of the other. 

It must not however be supposed that if a theorem is true, its 
converse is necessarily true. [See p. 25.] 



24 



GEOMETRY 



Theorem 7. [Euclid I. 8] 

If two triangles have the three sides of the one equal to the three 
sides of the other, each to each, they are equal in all respects. 
A D 





Let ABC, DEF be two triangles in which 

AB = DE, 

AC = DF, 

BC = EF. 
It is required to prove that the triangles are equal in all respects. 



Proof. 



because 



Apply the A ABC to the A DEF, 
so that B falls on E, and BC along EF, and 
so that A is on the side of EF opposite to D. 
Then because BC = EF,C must fall on F. 
Let GEF be the new position of the A ABC. 
Join DO. 
Because ED = EG, 
.'. the Z EDG = the Z EGD. Theor. 5. 

Again, because FD = FG, 
:. the Z FDG - the Z FGD. 
Hence the whole Z EDF = the whole Z A'GF ; 
that is, the Z EDF = the Z 5^C. 
Then in the A B AC, EDF ; 

BA = ED, and AC = DF, 

and the included Z BAC = the included Z ^Df' ; 

.'. the triangles are equal in all respects. Theor. 4. 

Q.E.D. 



CONGRUENT TRIANGLES 



25 



Obs. In this Theorem 

it is given that AB = DE, BC = EF, CA = FD; 

and we prove that ZC = IF, ZA= ZD, ZC = ZE. 
Also the triangles are equal in area. 

Notice that the angles which are proved equal in the two 
triangles are opposite to sides which were given equal. 

Note 1. We have taken the case in which DG falls within the 
A EDF, EGF. 

Two other cases might arise : 

(i) DG might fall outside the A EDF, EGF [as in Fig. 1.] 
(ii) DG might coincide with DF, FG [as in J'ig. 2.] 








These cases will arise only when the given triangles are obtuse- 
angled or right-angled ; and (as will be seen hereafter) not even then, 
if we begin by choosing for superposition the greatest side of the 
A ABC, as in the diagram of page 24. 

Note 2. Two triangles are said to be equiangular to one another 
when the angles of one are respectively equal to the angles of the 
other. 

Hence if two triangles have the three sides of one severally equal to 
the three sides of the other, the triangles are equiangular to one another. 

The student should state the converse theorem, and shew by a 
diagram that the converse is not necessarily true. 

*^,* At this stage Problems 1-5 and 8 [see page 70] may 
conveniently be taken, the proofs affording good illustrations of 
the Identical Equality of Two Triangles. 



26 GEOMETRY 



EXERCISES 



On the Identical Equality of Two Thiangles 
Theorems 4 and 7 

(Theoretical) 

"^1. Shew that the straight line which joins (lie vertex of an 
isosceles triangle to the middle point of the base, 

(i) bisects the vertical angle : (ii) is perpendicular to the base. 
/-^ 2. If A BCD is a rhombus, that is, an equilateral foursidcd 
figure; shew, by drawing the diagonals AC, BD, that 
(i) the angle ABC = the angle ADC; 
(ii) AC bisects each of the angles BAD, BCD. 
(iii) the diagonals bisect one another at right angles. 
.3. If in a quadrilateral A BCD the opposite sides are equal, 
namely AB = CD and AD = CB; prove that the angle ADC = 
the angle ABC. 

4. If ABC and DBC are two isosceles triangles drawn on the 
same base BC, prove (by means of Theorem 7) that the angle A B D 
= the angle ACD, taking (i) the case where the triangles are on the 
satne side of BC, (ii) the case where they are on opposile sides of BC. 
-Lj^ o. If ABC, DBC are two isosceles triangles drawn on opposite 
sides of the same base BC, and if A D be joined, prove that each of 
the angles BAC, BDC will be divided into two equal parts. 

6. Shew that the straight lines which join the extremities of the / 
/\base of an isosceles triangle to the middle points of the opposite ^^_v^ 
sides are equal to one another. 

^ 7. Two given points in the base of an isosceles triangle are equi- 
/oTstant from the extremities of the base; shew that they are also 
equidistant from th<^ vertex. n 

s^ 8. Shew tliat the triangle formed by joining the middle points 
of the sides of an equilateral triangle is also equilateral. 

9. ABC is an isosceles triangle having AB equal to AC; and/( 
the angles at B and C are bisected by BO and CO : shew that 

(i) BO = CO; (ii) .10 bisects the angle /?. IT. 

10. The equal sides BA, CA of an isosceles triangle HAC are 
pro(hiced beyond the vertex A to the points K and F, so that .1 E is 
equal to .1 /'' ; and FB, I'JC ant joined : shew that FB is equal to EC. 




TRIANGLES 27 

EXERCISES ON TRIANGLES 

( Numerical and Graphical) 

1. Draw a triangle ABC, having given a = 20", b = 2-1," 
c = r3". Measure the angles, and find their sum. 

2. In the triangle ABC, a = 7o em., b = 70 era., c = G'5 em. 
Draw and measure the perpendicular from B on CA. 

3. Draw a triangle ABC, in which a = 7 cm., fe = G cm., 
C = 65°. 

How would you prove theoretically that any two triangles having 
these parts are alike in size and shape? Invent some e.xperimental 
illustration. 

4. Draw a triangle from the following data: b = 2", c = 2'o", 
A = 57°; and measure a, B, and C. 

Draw a second triangle, using as data the values just found for a, 
B, C ; and measure b, c, A. What conclusion do you draw? 

5. When the sun is 42° alxtve the horizon, a vertical pole casts 
a -shadow 30 ft. long. Represent this on a diagram (scale 1" to 10 
ft.) ; and find by measiu'ement the approximate height of the pole. 

6. From a point .4 a surveyor goes 150 yards due East to B ; 
then 300 yards due North to C ; finally 450 yards due West to D. 
Plot his course (scale 1" to 100 yards) ; and find roughly how far D 
is from A. Measure the angle DAB, and say in what direction D 
bears from A. 

7. B and C are two points, knov.-n to be 260 yards apart, on a 
straight shore. A is a vessel at anchor. The angles CBA, BCA 
are observed to be 33° and 81° respectively. Find graphically the 
approximate distance of the vessel from the points B and C, and 
from the nearest point on shore. 

8. In surveying a park it is required to find the distance be- 
tween two points .4 and B ; but as a lake inter\'enes, a direct meas- 
urement cannot be made. The surveyor therefore takes a third 
point C, from which both 4 and B are accessible, and he finds CA 

= 245 yards, CB = 320 yards, and the angle ACB = 42°. Ascer- 
tain from a plan the approximate distance between A and B. 



28 GEOMETRY 

Theorem 8. [FAiclid I. 16] 

If one side of a triangle is produced, then the exterior angle is 
greater than either of the interior opposite angles. 




Let ABC be a triangle, and let BC be produced to D. 
It is required to prove that the exterior Z ACD is greater than 
either of the interior opposite A ABC, BAC. 

Suppose E to be the middle point of AC. 
Join BE ; and produce it to F, making EF equal to BE. 

Join FC. 
Proof. Then in the A AEB, CEF, 

{ AE = CE, 

because EB = EF, 

[and the Z AEB = the vertically opposite Z CEF ; 
/. the triangles are equal in all respects ; Theor. 4. 
so that the Z BAE = the Z ECF. 
But the Z ECD is greater than the Z ECF ; 
.'. the Z ECD is greater than the Z BAE ; 
that is, the Z ACD is greater than the Z BAC. 
In the same way, if AC is produced to (7, by supposing A 
to be joined to the middle point of BC, it may be proved that 
the Z BCG is greater than the Z ABC. 

But the Z BCG = the vertically opposite Z ACD. 
:. the Z ylCD is greater than the Z ABC. q.e.d. 



TRIANGLES 29 

Corollary 1. Any two angles of a triangle are together less 
than two right angles. 

A 

For the Z ABC is less than the Z ACD: Proved. 

to each add the Z ACB. 
Then the A ABC, ACB areleas than the^ ^CD, ACB, 
therefore, less than two right angles. 

BCD 

Corollary 2. Every triangle must have at least two acute 
angles. 

For if one angle is obtuse or a right angle, then by Cor. 1 each of 
the other angles must be U'ss than a right angle. 

Corollary 3. Only one perpendicular can be drawn to a 
straight line from a given point outside it. 




If two perpendiculars could be drawn to .4 B from 
P, we should have a triangle PQR in which each of 
the A PQR, PRQ would be a right angle, wiiich is 
impossible. 



EXERCISES 



P 



A Q R B 



1. Prove Corollary 1 by joining the vertex A to any point in the 
base BC. 

2. ABC is a triangle and D any point within it. If BD and CD 
are joined, the angle BDC is greater than the angle BAC. Prove this 

(i) by producing BD to meet AC. 
(ii) by joining .4 D, and producing it towards the base. 

3. If any side of a triangle is produced both ways, the exterior 
angles so formed are together greater than two right angles. 

4. To a given straight line there cannot be drawn from a point 
outside it more than two straight lines of the same given length. 

5. If the equal sides of an isosceles triangle are produced, the 
exterior angles must be obtuse. 



30 GEOMETRY 

Theorem 9. [Euclid I. 18] 

// one side of a triangle is greater than another, then the angle 
opposite to the greater side is greater than the angle opposite to 
the less. 

A 




Let ABC be a triangle, in which tlie side AC is greater than 
the side AB. 

It is required to prove that the Z ABC is greater than the 
Z ACB. 

From AC cut off AD equal to AB. 
Join BD. 

Proof. Because A B = AD, 

.: the Z ABD = the Z ADB. Theor. 5. 

But the exterior Z ADB of the A BDC is greater than the 
interior opposite Z DCB; that is, greater than the Z ACB. 

.: the Z ABD is greater than the Z ACB. 
Still more then is the Z ABC greater tlian the Z ACB. 

Q.E.D. 

Ohs. The mode of domonstration used in the following Theorem 
is known as the Proof by Exhaustion. It is applicable to eases in 
whieh one of eertain snpponitions must necessarily be true; and it 
consists in shewinfj; that each of these su[)positions is false /////( our 
exception: hence the truth of the remaining supposition is inferred. 



INEQUALITIES 31 



Theorem 10. [Euclid I. 19] 

// one angle of a triangle is greater than another, then the side 
opposite to the greater angle is greater than the side opposite to 
the less. 

A 




Lot ABC be a ti-ianglc, in which the Z ABC is greater than 
the Z ACB. 

It is required to prove that the side AC is greater than the 
side AB. 

Proof. If AC is not greater than AB, 

it must be either equal to, or less than AB. 
Now if AC were equal to AB, 
then the Z ABC would be equal to the Z ACB ; Thcor. o. 
but, by hypothesis, it is not. 

Again, if AC were less than AB, 
then the Z A 5C would be less than the Z ACB ; Theor.d. 
but, by hypothesis, it is not. 

That is, AC is neither equa^to, nor less than AB. 

.'. AC is greater than AB. q.e.d. 

[For Exercises on Theorems 9 and 10 see page 34.] 



32 GEOMETRY 

Theorem U. [Euclid I. 20] 

A ny two sides of a triangle arc together greater than the third 

side. 

D 




Let ABC be a triangle. 
It is required to prove that any two of its sides are together 
greater than the third side. 

It is enough to shew that if BC is the greatest side, then 
BA, AC are together greater than BC. 

Produce BA to D, making AD equal to AC. 
Join DC. 

Proof. Because AD = AC, 

:. the Z AC D = the Z ADC. Theor. 5. 
But the Z BCD is greater than the Z ACD ■ 
.: the Z BCD is greater than the Z ADC, 
that is, than the Z BDC. 
Hence from the A BDC, 

BD is greater than BC. Theor. 10. 

But BD = BA and AC together ; 
/. BA and AC are together greater than BC. 

Q.E.D. 

Note. This proof may serve as an exorcise, but the truth of the 
Tlieoreni is really self-evident. For to so from B to (' along the 
straight line liC is elearly shorter lliaii to i^o from li (o .1 and (hen 
from A to C Jn other words 

The shortest distance between two points is the straight line which 
joins them. 



INEQUALITIES 33 

Theorem 12 

Of all straight lines draicn from a given point to a given 
straight line the perpendicular is the least. 




A R Q C P B 

Let OC be the perpendicular, and OP any oblique, drawn 
from the given point to the given straight line AB. 
It is required to prove that OC is less than OP. 
Proof. In the A OCP, since the Z OCP is a right angle, 
.*. the Z OPC is less than a right angle ; Theor. 8. Cor. 
that is, the Z OPC is less than the Z OCP. 

:. OC is less than OP. Theor. 10. 

Q.E.D. 

Corollary 1. Hence conversely, since there can be only 
one perpendicular and one shortest line from to AB, 

If OC is the shortest straight line from to AB, then OC is 
perpendicular to AB. 

Corollary 2. Two obliques OP, OQ, which cut AB at equal 
distances from C, the foot of the perpendicular , are equal. 

The A OCP, OCQ may be shewn to be congruent by Theorem 4 ; 
hence OP ^=^0Q. 

Corollary 3. Of two obliques OQ, OR, if OR cuts AB 
at the greater distance from C, the foot of the perpendicular, then 
OR is greater than OQ. 

The Z OQC is acute, .-. the Z OQR is obtuse; 

.-. the Z OQR is greater than the Z ORQ; 
:. OR is greater than OQ. 
D 




34 GEOMETRY 

\ 
EXERCISES OX INEQUALITIES IX A TRIAXGLE 

Xl. The hypotenuse is the greatest side of a right-angled triangle. 
X 2. The greatest side of any triangle makes acute angles with each 

k^^/ the other sides. 
"^ 3. If from the ends of a side of a triangle, two straight lines are 

^ drawn to a point within the triangle, then these straight lines are together 
>^^y/c6s than the other two sides of the triangle. 

. 4. BC, the base of an isosceles triangle ABC, is produced to any 
point D ; shew that A D is greater than either of the equal sides. > 

_. 5. If in a quadrilateral the greatest and least sides are opposite 
tIo one another, then each of the angles adjacent to the least side is 
greater than its opposite angle. 

(). In a triangle ABC, if AC is not greater than AB, shew that 
,%v any straight line drawn through the vertex A and terminated by the ^ 

^f^ base BC, is less than AB. h 

^ 7. ABC is a triangle, in which OB, DC bisect the angles ABC, 

'ACB respectively : shew that, if A B is greater than AC, then OB isH 
'(_, 'greater than OC. 

The difference of any two sides of a triangle is less than the f 
third side. 
■SlyJirS.- The sum of the distances of any point from the three angular i 
points of a triangle is greater than half its perimeter. S~ 

' ^v 10. A BC is a triangle, and the vertical angle BAC is bisected by 
«- line which meets BC in A' ; shew that BA is greater than BX, and 
CA greater than CX. Hence obtain a proof of Theorem 11. /n 1 

11. The sum of the distances of any point within a triangle 
from its angular points is less than the perimeter of the trianglo^ 

12. The sum of the diagonals of a quadrilateral is not greater 
tlian the sum of the four straight lines drawn from the angular points 
to any given point. In what case are these sums equal? 

13. In a triangle any two sides are together greater than twice the 
median which hiserls the remaining side. ' ,'/.'; 

[Produce the median, and complete the construction after the 
manner of Theorem S.] 
_jL 14. In any triangle the sum of the medians is less than, the perim- 
iXcr. ^ 

^ . ho 




L 



PARALLELS 35 



PARALLELS i"-}- 

Definition. Parallel straight lines are such as, being in 
the same plane, do not meet however far they are produced 
beyond both ends. 

Note. Parallel lines must be in the same plane. For instance, 
two straight lines, one of which is drawn on a table and the other on 
the floor, would never meet if produced ; but they are not for that 
reason necessarily parallel. 

Axiom. Two intersecting straight lines cannot both be paral- 
lel to a third straight line. 

In other words : 

Through a given point there can be only one straight line 
parallel to a given straight line. 

This assumption is known as Playfair's Axiom. 

Definition. When two straight Hnes AB, CD are met by 
a third straight hne EF, eight angles are formed, to which 
for the sake of distinction particular names are given. 

Thus in the adjoining figure, 
1, 2 7, 8 are called exterior angles, 
3, 4, 5, 6 are called interior angles, 
4 and 6 are said to be alternate angles ; 
so also the angles 3 and 5 are alternate 
to one another. — ^ 

Of the angles 2 and 6, 2 is referred 
to as the exterior angle, and 6 as the 
interior opposite angle on the same side 
of EF. Such angles are also known as corresponding angles. 
Similarly 7 and 3, 8 and 4, 1 and 5 are pairs of corresponding 
angles. 




36 GEOMETRY 

Theorem 13. [Euclid I. 27 and 28] 

If a straight line cuts two other straight lines so as to make 
(i) the alternate angles equal, 
or (ii) an exterior angle equal to the interior opposite angle on 

the same side of the cutting Ihu, 
or (iii) the interior angles on the same side equal to two right 

angles ; 
then in each case the two straight lines are parallel. 




(i) Let the straight line EGHF cut the two straight Hues 
AB, CD at G and H so as to make the alternate A AG II, 
GHD equal to one another. 

It is required to prove that AB and CD are parallel. 

Proof, li AB and CD are not parallel, they will meet, if 
produced, either towards B and D, or towards A and C. 
If possible, let AB and CD, when produced, meet towards B 

and D, at the point K. 
Then KGH is a triangle, of which one side KG is produced 

to^ ; 
.'. the exterior Z AGH is greater than the interior opposite 

Z GHK ; but, by hypothesis, it is not greater. 
.". AB ixmX CD cannot meet when produced towards B an<l />. 
Similarly it may be shewn that tluy cannot meet towards 

A and C : 

.'. AB and CD are parallel. 




37 



(ii) Let the exterior Z EGB = the interior opposite 
Z GHD. 

It is required to prove that AB and CD are parallel. 
Proof. Because the Z EGB = the Z GHD, ^ 

and the Z EGB = the vertically opposite Z ^ Gi/ ; ' Tt* r 

.-. the Z .46'// = the Z G'//D : jQ] 

and these are alternate angles ; 
.". AB and CD are parallel. 
(iii) Let the two interior A BGH, GHD be together equal 
to two right angles. 
It is required to prove that AB and CD are parallel. 
Proof. Because the A BGH, GHD together = two right 
angles ; 

and because the adjacent A BGH, A GH together = two right 
angles ; 

/. the A BGH, AGH together = A BGH, GHD. 

From these equals take the Z BGH ; 

then the remaining Z AGH =~the remaining Z GHD : 

and these are alternate angles ; 

.'. AB and CD are parallel. q.e.d. 

Definition. A straight line drawn across a set of given 
lines is called a transversal. 

For instance, in the abo^'e diagi-am the line EGHF, which crosses 
the given Unes AB, CD, is a transversal. 



38 GEUMKTllY 

Theorem 14. [Euclid I. 29] 

7/ a straight line cuts two parallel lines, it makes 
(i) the alternate angles equal to one another ; 

(ii) the exterior angle equal to the interior opposite angle on 
the same side of the cutting line ; 

(iii) the two interior angles nn the same side together equal to 
two right angles. 

rE 




Let the straight hues AB, CD be parallel, and let the 
straight line EGHF cut them. 
It is required to prove that 
(i) the Z AGH = the alternate Z OHD ; 
(ii) the exterior Z EOB = the interior opposite Z (ilU) ; 
(iii) the two interior A BGH, GHD together = two right 
angles. 

Proof, (i) If the Z AGH is not equal to the Z GHD, 
suppose the Z P(f'// equal to the Z GHD, and alternate to it ; 
then PG and CD are parallel. Theor. 13. 

But, by hypothesis, .47? and CD are i)arallel ; 
.'. the two intersecting straight lines AG, PG are both i)arallcl 
to CD : which is impossible. Plaiifair\ Axiom. 

:. the Z AGH is not une(]ual to the Z GHD ; 
that is, the alternate A AGH, GHD are equal. 



PARALLELS 39 

(ii) Again, Ijccausc the Z E(W = the vertically opposite 
Z -AGH ; 

and the Z AGH = the alternate Z GHD; Proved. 

.'. the exterior Z JS'CrB = the interior opposite Z GHD. 

(iii) Lastly, the Z EGB = the Z 6'///) ; Proved. 

add to each the Z fif/H ; 

then the A EGB, BGH together = the angles BGH, GHD. 

But the adjacent A EGB, BGH together = two right 
angles ; 

.*. the two interior A BGH, GHD together = two right 
angles. q.e.d. 

PARALLELS ILLUSTRATED BY ROTATION 

The direction of a straight line is determined by the angle which 
it makes \vith some given line of reference. 

Thus the direction of AB, relatively to the given lino Y'X, is given 
by the angle A PX. 

Now suppose that AB and CD in 
the adjoining diagram are paral- 
lel ; then we have learned that the 
ext. Z APX = the int. opp. Z CQX ; 
that is, AB and CD make equal 
angles with the line of reference 
YX. 

This brings us to the leading idea 
connected with parallels : 
Parallel straight lines have the same direction, but differ in position. 

The same idea may be illustrated thus : 

Suppose AB to rotate about P through the Z APX, so as to 
take the position AT. Thence let itlfotate about Q the opposite way 
through the equal Z XQC : it will now take the position CD. Thus 
A B may be brought into the position of CD by two rotations which, 
being equal and opposite, involve no final change of direction. 

Ohs. If ^B is a straight line, movements from A towards 
B, and from B towards A are said to be in opposite senses 
of the line AB. 




40 GEOMETRY 

Theorem 15. [Euclid I. 30] 

Straight lines which are parallel to the same straight line are 
parallel to one another. 

/ 

A G/ B 



HA 



K4' Q 



Let the straight hnes AB, CD be each parallel to the straight 
line PQ. 

It is required to prove that AB and CD are parallel to one 
another. 

Draw a straight line EF cutting AB, CD, and PQ in the 
points G, H, and K. 

Proof. Then because AB and PQ are parallel, and EF 
meets them, 

.-. the Z AGK = the alternate Z GKQ. 
And because CD and PQ are parallel, and EF meets them, 
.'. the exterior Z GHD = the interior opposite Z GKQ. 
.: the Z AGH = the Z GHD ; 
and these are alternate angles ; 

.*. AB and CD are parallel. q.e.d. 

Hypothetical Construction. In the diagram on p. 39 
let AB be a fixed straight line, Q a fixed point, CD a straight 
line turning about Q, and YQPX any transversal through Q. 
Then as CD rotates, there nmst be one position in which the 
Z CQX = the fixed Z APX. 

Hence through any given point we may assume a line to pass 
parallel to any given straight line. 



PARALLELS 41 



EXERCISES ON PARALLELS 

1. In the diagram of the previous page, if the angle EGB is 55°, 
express in degrees each of the angles GHC, HKQ, QKF. 

_L_ 2. Straight lines which are perpendicular to the same straight line 
are parallel to one another. 

_;_ 3. If a straight line meets two or more parallel straight lines, and is 
perpendicular to one of them, it is also perpendicular to all the others. 

^>j 4. Angles of which the arms are parallel, each to each, are either . 
ywttaZ or supplementary. 

5. Two straight lines AB, CD bisect one another at 0. Shew 
that the straight lines joining AC and BD are parallel. S" 

6. Any straight line drawn parallel to the base of an isosceles 
triangle makes equal angles with, the sides. 

V 7. If from any point in the bisector of an angle a straight line is 
drawn parallel to either arm of the angle, the triangle thus formed is 
isosceles. 

)§^ From X, a point in the base BC of an isosceles triangle ABC, 
straight line is drawn at right angles to the base, cutting AB in J', -J 



7^ 



and CA produced in Z: shew the triangle AYZ is isosceles. 

9. If the straight line which bisects an exterior angle of a tri- / 
angle is parallel to the opposite side, shew that the triangle is / ; 
isosceles. Sv p ' 

10. The straight lines drawn from any point in the bisector of 
an angle parallel to the arms of the angle, and terminated by them, 
are equal ; and the resulting figure is a rhombus. 

11. AB and CD are two straight lines intersecting at D, and the 

adjacent angles so formed are bisected : if through any point A' in _J 

DC a straight line fXZ is drawn parallel to AB and meeting the bi- ' 
sectors in Y and Z, shew that Z F is equal to XZ. 

12. Two straight rods PA, QB revolve about pivots at P and Q, 
PA making 12 complete revolutions a minute, and QB making 10. 
If they start patrallel and pointing the same way, how long wiU it be 
before they are again paralfel, (i) pointing opposite ways, (ii) point- 
ing the same way? a 




42 GEOMETRY 

Theorem 16. [Euclid I. 32] 

The three angles of a triangle are together equal to two right 
angles. 




B CD 

Let ABC be a triangle. 
It is required to prove that the three A ABC, BCA, CAB 
together = two right angles. 

Produce BC to any point D ; and suppose CE to be the 
line through C parallel to BA. . 

Proof. Because BA and CE are parallel and ^4^ meets 
them, 

.-. the Z ACE = the alternate Z CAB. 

Again, because BA and CE are parallel, and BD meets 
them, 

.*. the exterior Z ECD = the interior opposite Z ABC. 
.'. the ivhole exterior Z ACD = the sum of the two interior 
opposite A CAB, ABC. 

To each of these equals add the Z BCA ; then 
the A 5CA, A CD together = the three A BCA, CAB, ABC. 
But the adjacent A BCA, ACD together = two right 
angles. 

.'. the A BCA, CAB, ABC together = two right angles. 

Q.E.D. 

Ohs. In the course of this proof the following most im- 
portant property has been established. 

// a side of a triangle is produced, the exterior angle is equal to 
the sum of the two interior opposite angles. 

Namely, the ext. Z ACD = the Z CAB + the Z ABC. 




THE ANGLES OF A TRIANGLE 43 

6 

INFERENCES FROM THEOREM 16 

1. If A, B, and C denote the number of degrees in the angles 
of a triangle, 

then A + B + C = 180°. 

2. If two triangles have two angles of the one respectively 
equal to two angles of the other, then the third angle of the one is 
equal to the third angle of the other. 

3. hi any right-angled triangle the two acute angles are com- 
plementary. 

4. If one angle of a triangle is equal to the sum of the other 
two, the triangle is right-angled. 

5. The sum of the angles of any quadrilateral figure is equal 
to four right angles. 

EXERCISES ON THEOREM 16 

1. Each angle of an equilateral triangle is 60°. 

2. In a right-angled isosceles triangle the angles are 45°, 45°, 90°. 

_^ 3. Two angles of a triangle are 36° and 123° respectively : de- 
^auce the third angle ; and verify your result by measurement. • i ' 

4. In a triangle ABC, the Z B = 111°, the Z C = 42°; de- 
duce the Z A, and verify by measurement. 

5. One side BC of a triangle ABC is produced to D. If the 
exterior angle ACD is 134°, and the angle B AC is 42°, find each of 
the remaining interior angles. -i. c 

6. In the figure of Theorem 16, if the Z ACD = 118°, and the 
/. B = 51°, find the A A and C ; and check your results by measure- 
ment. 

7. Prove that A + B + C = 180° by supposing a line drawn through^J 
the vertex parallel to the base. 

8. // two straight lines are perpendicular to two other straight lines, 

•, each to each, the acute angle betiveen the first pair is equal to the acute "-^ 
/^n^ie between the second pair. 



44 GEOMETRY 

Corollary 1. All the interior angles of any rectilineal 
figure, together with four right angles, are equal to twice as many 
right angles as the figure has sides. 

D 




Let ABODE be a rectilineal figure of n sides. 

It is required to prove that 

all the interior angles + 4 rt. A = 27i rt. A . 
Take any point within the figure, and join to each of 
its vertices. 

Then the figure is divided into n triangles. 
And the three A of each A together = 2 rt. A . 
Hence all the A of all the A together = 2n rt. A . 
But all the A of all the A make up all the interior angles 
of the figure together with the angles at 0, which = 4 rt. A . 
.: all the int. A of the figure + 4 rt. ^4 = 2n rt. A . 

Q.E.D. 

Definition. A regular polygon is one which has all its 
sides equal and all its angles equal. 

Thus if D denotes the number of tU>grces in each angle of 
a regular polygon of n sides, the above result may be stated 
thus : 

nD + 300° = n • 180°. 

EXAMPLE 

Find the number of degrees in each angle of a regular 
(i) hexagon (6 sides) ; (ii) octagon (8 sides) ; (iii) decagon 
(10 sides). 



THE ANGLES OF RECTILINEAL FIGURES 45 

EXERCISES ON THEOREM 16 

( Numerical and Graphical) 

'* 1. ABC is a triangle in which the angles at B and C are re- 
spectively double and treble of the angle at A : find the number of 
degrees in each of these angles. i) 

y 2. The base of a triangle is produced both ways, and the exterior ^* 
angles are found to be 94° and 126° ; deduce the vertical angle. 
Construct such a triangle, and check your result by measurement. 

/' 3. The sum of the angles at the base of a triangle is 162°, and "K 
their difference is 60° : find all the angles. 

4. The angles at the base of a triangle are 84° and 62° ; deduce 
(i) the vertical angle, (ii) the angle between the bisectors of the base - 
angles. Check your results by construction and measurement. ""^f^ 

5. In a triangle A EC, the angles at B and C are 74° and 62° ; if 
AB aAid AC are produced, deduce the angle between the bisectors of 
the exterior angles. Check your result graphically. 

6. Three angles of a quadrilateral are respectively 1145°, 50°, 
and 755°; find the fourth angle. 

7. In a quadrilateral A BCD, the angles at B, C, and D are re- 
spectively equal to 2 A, 3 A, and 4 A ; find all the angles. 

8. Four angles of an irregular pentagon (5 sides) are 40°, 78°, 
122°, and 135° ; find the fifth angle. / t' ^ 

9. In any regular polygon of n 'sides, each angle contains 

^ '""'•'-' right angles. \\^ 'X 

n ^ '^, 

(i) Deduce this result from the Enunciation of Corollary 1. 
(ii) Prove it independently by joining one vertex A to each of 
the others (except the two immediately adjacent to A), thus divid- 
ing the polygon into ii-2 triangles. 

10. How many sides have the regular polygons each of whoSte/ 
angles is (i) 108°, (ii) 156°? ^ 

11. Shew that the only regular figures which may be fitted to- 
gether so as to form a plane surface are (i) equilateral triangles, (ii) 
squares, (iii) regular hexagons. 



46 



GEOMETRY 



Corollary 2. // the sides of a rectilineal figure, which has 
no re-entrant angle, are produced in order, then all the exterior 
angles so formed are together equal to four right angles. 




1st Proof. Suppose, as before, that the figure has n sides. 
Now at each vertex' 

the interior Z + the exterior Z = 2 rt. A . 
.'. the sum of the n int. A + the sum of the n ext. A = 2n 
rt. A. 

But by Corollary 1, 

the sum of the int. A -\- 4 rt. ^ = 2/i rt. Z ; 

.". the sum of the ext. Z = 4 rt. Z . 
2nd Proof. 



Q.E.D. 





Take any point 0, and suppose Oa, Ob, Oc, Od, and Oe are 
lines parallel to the sides marked A, B, C, D, E (and drawn 
from in the .sense in which those sides were produced). 

Then the ext. Z between the sides A and B = the Z aOb. 

The other ext. A = the respective A bOc, cOd, dOe, eOa. 

.'. the sum of the ext. A = the sum of the Z at 

= 4 rt. A . 



THE ANGLES OF RECTILINEAL FIGURES 



47 



EXERCISES 

I. If one side of a regular hexagon is produced, shew that tiie 
exterior angle is equal to the interior angle of an equilateral triangle. 
, 2. Express in degrees the magnitude of each exterior angle of 
y(i)& regular octagon, (ii) a regular decagon. Q (- '' 

3. How many sides has a regular polygon if each exterior angle 
is (i) 30°, (ii) 24°? i< 

4. If a str3,ight line meets two parallel straight lines, and the two 
interior angles on the same side are bisected, shew that the bisectors 
meet at right angles. 

5. If the base of any triangle is produced both ways, shew that 
the sum of the two exterior angles mirius the vertical angle is equal 
to two right angles. 

6. In a triangle ABC the base angles at B and C are bisected 
by BO and CO respectively. Shew that the angle BOC = 90° + ^ . 

7. In the triangle A BC, the sides AB, AC are produced, and the 
exterior angles are bisected by BO and CO. Shew that the angle 

A 
2 ■ 

8. The angle contained by the bisectors of two adjacent angles 
of a quadrilateral is equal to half the sum of the remaining angles. 

9. The straight hne joining the middle point of the hypotenuse 
of a right-angled triangle to the right angle is equal to half the 
hypotenuse. 



BOC 



90° - 



i'i> 



EXPERIMENTAL PROOF OF THEOREM 16 
[A + B + C = 180°] 

In the A ABC, AD is perp. to BC, the 
greatest side. AD is bisected at right 
angles by ZF ; and YP, ZQ are perps. on 
BC. 

If now the A is folded about the three 
dotted lines, the A A, B, and C will coin- 
cide with the AZDY,ZDQ, YDP; 
.: their sum is 180°. 




D P C 



48 GEOMETRY 



Theorem 17. [Euclid T. 26] 

If two triangles have two angles of one equal to two angles of 
the other, each to each, and any side of the first equal to the cor- 
responding side of the other, the triangles are equal in all respects. 





Let ABC, DEF be two triangles in which 
the I A = the Z D, 
the Z 5 = the Z E, 
and the side BC = the corresponding side EF. 

It is required to prove that the A ABC, DEF are equal in all 
respects. 

Proof. The sura of the A A, B, C = 2 rt. A Theor. IG. 
= the sum of the A D, E, and F ; 
and the A A and B = the A D and E respectively, 
.-. the Z C = the Z F. 
Apply the A ABC to the A DEF, so that B falls on E, 
and BC along EF. 

Then, because BC = EF, C must coincide with F. 
Because the A B = the Z E, BA must fall along ED. 
And because the A C = the Z F, CA must fall along FD. 
.'. the point A, which falls both on ED and on FD, must coin- 
cide with D, the point in which these lines intersect. 
.-. the A ABC coincides with the A DEF, 
and is therefore equal to it in all respects. 
So that AB = DE, and AC = DF; 
and the A DBC = the A DEF in area. q.e.d. 



CONGRUENT TRIANGLES 49 

EXERCISES 
On the Identical Equality of Triangles 




t 




1. Shew that the perpendiculars drawn from the extremities of e>.^0 
the base of an isosceles triangle to the opposite sides are equal. ' 

2. Any point on the bisecloi- of an angle is equidistant from the 
arms of the angle. ^ 

3. Through 0, the middle point of a straight line AB, any // 
straight line is drawn, and perpendiculars AX and BY are droppedy k > >Vm}^ 
upon it from A and B: shew that AX is equal to BY. %^ 

4. If the bisector of the vertical angle of a triangle is at right *" 
''( angles to the base, the triangle is isosceles. 

5. If in a triangle the perpendicular from the vertex on the base / A\*H 



bisects the base, then the triangle is isosceles. 

a" 
^ If the bisector of the vertical angle of a triangle also bisects 
the base, the triangle is isosceles. 

[Produce the bisector, and complete the construction after the 
manner of Theorem 8.] 

7. The middle point of any straight line which meets two 
parallel straight lines, and is terminated by them, is equidistant 
from the parallels. 

8. A straight line drawn between two parallels, and terminated 
by them, is bisected ; shew that any other straight line passing 
through the middle point and terminated by the parallels is also 
bisected at that point. 

9. If through a point equidistant from two parallel straight 
lines, two straight lines are drawn cutting the parallels, the portions 
of the latter thus intercepted are equal. 

^ 10. A surveyor wishes to ascertain the breadth of a river which he 
'cannot cross. Standing at a point A near the bank, he notes an object 
B immediately opposite on the other bank. He lays down a line AC of 
any length at right angles to AB, fixing a mark at 0, the middle point of 
AC. From C he walks along a line perpendicular to AC until he 
reaches a point D from which O and B are seen in the same direction. 
He noio measures CD : prove that the result gives him the width of the 
river. 

E 



¥ 



50 GEOMETRY 

ON THE IDENTICAL EQUALITY OF TRIANGLES 

Three cases of the congruence of triangles have been dealt 
with in Theorems 4, 7, 17, the results of which arc : 

Two triangles are equal in all respects when the following 
three parts in each are severally equal : 

1. Two sides, and the included angle. Theorem 4. 

2. The three sides. Theorem 7. 

3. Two angles and one side, the side given in one triangle 
CORRESPONDING to that given in the other. Theorem 17. 

Two triangles are not, however, necessarily equal in all 
respects when any three parts of one are equal to the corre- 
sponding parts of the other. 

For example : 

(i) When the three angles of one are 
equal to the three angles of the othei", 
each to each, the adjoining diagram 
shews that the triangles need not be 
equal in all respects. 

(ii) When two sides and one angle in one are equal to two 
sides and one angle of the other, the given angles being opposite 
to equal sides, the diagram below shews that the triangles 
need not be equal in all respects, 

>^ D 





B C E p- F 

For ii AB = DE, and AC = DF, and the Z ABC = the 
Z DEF, it will be seen that the shorter of the given sides in 
the triangle DEF may lie in either of the positions DF or DF'. 

Note. See also Theorem IS, p. 51, and Problem 0, p. 85. 



CONGRUENT TRIANGLES 



51 



TllEOKEM 18 

Two right-angled triangles which have their hypotenuses equal, 
and one side of one equal to one side of the other, are equal in all 
respects. 





Let ABC, DEF be two right-angled triangles, in which 
the A ABC, DEF are right angles, 
the hypotenuse AC = the hypotenuse DF, 
and AB = DE. 
It is required to prove that the A ABC, DEF are equal in all 
respects. 

Proof. Apply the A ABC to the A DEF, so that AB falls 
on the equal line DE, and C on the side of DE opposite to F. 
Let C be the point on which C falls. 
Then DEC represents the A ABC in its new position. 
Since each of the A DEF, DEC is a right angle, 

/. EF and EC are in one straight line. 
And in the A CDF, because DF = DC {i.e. AC), 

.: the Z DEC = the Z DCF. Theor. 5. 
Hence in the A DEF, DEC, 
[ the Z DEF = the Z DEC, being right angles ; 
because] the Z DFE = the Z DCE, Proved 

[ and the side DE is common. 
/. the A DEF, DEC are equal in all respects ; Theor. 17. 
that is, the A DEF, ABC are equal in all respects. 

Q.E.D. 



52 



GEOMETRY 



♦Theorem 19. [Euclid I. 24] 

// livo triangles have two sides of the one equal to two sides of 
the other, each to each, but the angle included by the two sides of 
one greater than the angle included by the corresponding sides 
of the other ; then the base of that which has the greater angle is 
greater than the base of the other. 




KG 



Let ABC, DEF be two triangles, in which 

BA = ED, and AC = DF, 
but the Z BAC is greater than the Z EDF. 
It is required to prove that BC is greater than EF. 

Proof. Apply the A ABC to the A DEF, so that A falls 
on D, and AB along DE. 

Then because AB = DE, B must coincide with E. 
Let DG, GE represent AC, CB in their new position. 
Then if EG passes through F (Fig. 1), EG is greater than 
EF; that is, BC is greater than Ef. 

But if EG does not pass througV^ (Fig. 2), suppose that 
DK bisects the Z FDG, and meets EG in A'. Join FK. 
Then "in the A FDK, GDK, 
J FD = GD, and DK is common to both, 
becaus(> j^j^^j ^^^ included Z FDK = the included Z GDK; 
.: FK = GK. Thcor. 4. 

Now the two sides E K, KF are greater than EF; 
that is, EK, KG are greater than EF. 
.: EG (or BC) is greater than EF. q.e.d. 



CONVERSE OF THEOREM 19 



53 



Conversely, // two triamjles have two sides of the one equal to 
two sides of the other, each to each, but the base of one greater 
than the base of the other; then the angle contained by the sides 
of that which has the greater base is greater than the angle con- 
tained by the corresponding sides of the other. 

A D 




Let ABC, DEF be two triangles in which 
BA = ED, 
and AC = DF, 
but the base EC is greater than the base EF. 
It is required to prove that the Z BAC is greater than the 
/.EDF. 

Proof. If the Z BAC h not greater than the Z EDF, 
it must be either equal to, or less than the Z EDF. 
Now if the ^ BAC were equal to the Z EDF, 
then the base BC would be equal to the base EF ; Theor. 4. 
but, by hypothesis, BC is not equal to EF. 
Again, if the Z BAC were less than the Z EDF, 
then the base BC would be less than the base EF ; Theor. 19. 

but, by hypothesis, BC is not less than EF. 
That is, the Z BAC is neither equal to, nor less than the 
Z EDF ; 

.-. the Z BACis greater than the Z EDF. 

Q.E.D. 

* Theorems marked with an asterisk may be omitted or postponed 
at the discretion of the teacher. 



54 GEOMETRY 

REVISION LESSON ON TRIANGLES 

L State the properties of a triangle relating to 
(i) the sum of its interior angles ; 
(ii) the sum of its exterior angles. 
What property corresponds to (i) in a polygon of n sides? With 
what other figures does a triangle share the property (ii) ? 

2. Classify triangles with regard to their angles. Enunciate 
any Theorem or Corollary assumed in the classification. 

3. Enunciate two Theorems in which from data relating to the 
sides a conclusion is drawn relating to the angles. 

In the triangle ABC, if a = 3'6 cm., b = 28 cm., c = 3-6 cm., 
arrange the angles in order of their sizes (before measurement) ; and 
prove that the triangle is acute-angled. 

4. Enunciate two Theorems in which from data relating to the 
angles a conclusion is drawn relating to the sides. 

In the triangle ABC, if 

(i) A = 48° and B = 51°, find the third angle, and name the 
greatest side. 

(ii) A = B — 62 1 °, find the third angle, and arrange the sides 
in order of their lengths. 

5. From which of the conditions given below may we conclude 
that the triangles ABC, A'B'C are identically equal? Point out 
where ambiguity arises ; and draw the triangle A BC in each case. 

(^=^'=71°. fo=a'=4-2cm. f^=yl'=3G°. 

(i) j B = B' =46°. (ii) \h=h' =24 cm. (iii) \b = B' = 121°. 

I a =a'. =3-7 cm. I C=C' =81°. I C = r =23°. 

f a= a' =30 cm. 

(iv) b= 6' =5-2 cm. (v) 

I c = c' =4"5 cm. 

6. Summarise the results of the last question by stating gen- 
erally under what conditions two triangles 

(i) are necessarily congruent; 
(ii) may or may not be congruent.* 

7. If two triangles have their angles equal, each to each, the triangles 
are not necessarily equal in all respects, because the three data are not 
independent. Carefully explain this statement. 



R = /?'=53°. 




C=C' =90°. 


b = b' =43 cm. 


(vi) 


c= c' =5 cm. 


c= c' =5'0 cm. 




a= a' ='S cm. 



EXERCISES ON TRIANGLES 



(Miscellaneous Examples) 

8. (i) The perpendicular is the shortest line that can be drawn to a 
given straight line from a given point. 

(ii) Obliques which make equal angles with the perpendicular are 
equal. 

(iii) Of two obliques the less is that which ynakcs the smaller angle 
with the perpendicular. 

'^■" // two triangles have two sides of the one equal to two sides of 
the other, each to each, and have likewise the angles opposite to one pair 
of equal sides equal, then the angles opposite to the other pair of equal 
sides are either equal or supplementary, and in the former case the 
triangles are equal in all respects. 

10. PQ is a perpendicular (4 em. in length) to a straight line 
AT. Draw through P a series of obliques making with PQ the 
angles lo", 30°, 45°, G0°, 75°. Measure the lengths of these obliques, 
and tabulate the results. 

11. PAB is a triangle in which AB and AP have constant 
lengths 4 cm. and 3 cm. If AB in fixed, and AP rotates about A, 
trace the changes in PB, as the angle .4 increases from 0° to 180°. 

Answer this question by drawing a series of figures, increasing A 
by increments of 30°. Measure PB in each case, and tabulate the 
results. 

12. From B, the foot of a flagstaff A B, a horizontal hne is drawn, 
passing two points C and D which are 27 feet apart. The angles 
BCA and BDA are 65° and 40° respectively. Represent this on a 
diagram (scale 1 cm. to 10 ft.), and find bj^ measurement the ap- 
proximate height of the flagstaff. 

13. From P, the top of a lighthouse PQ, two boats A and B are 
seen at anchor in a line due south of the hghthouse. It is known 
that PQ = 126 ft., Z PAQ = 57°^ Z PBQ = .33°; hence draw a 
plan in which 1" represents 100 ft., and find by measurement the 
distance between A and B to the nearest foot. 

14. From a lighthouse L two ships A and B, which are 600 
yards apart, are observed in directions S.W. and 15° East of South 
respectively. At the same time B is observed from A in a S.E. 
direction. Draw a plan (scale 1" to 200 yds.), and find by measure- 
ment the distance of the lighthouse from each ship. 



56 



GEOMETRY 



PARALLELOGRAMS 

DEFINITIONS 

1. A quadrilateral is a plane figure bounded 
by four straight lines. 

The straight line which joins opposite angular 
points in a quadrilateral is called a diagonal. 




2, A parallelogram is a quadrilateral 
whose opposite sides are parallel. 

[It will be proved hereafter that the opposite 
sides of a parallelogram are equal, and that its 
opposite angles are equal.] 



3. A rectangle is a parallelogram whicli 
has one of its angles a right angle. 

[It will be proved hereafter that all the angles of 
a rectangle are right angles. See page 59.] 



4. A square is a rectangle which has two 
adjacent sides equal. 

[ 1 1 will be proved that all the sides of a square are 
equal and all its angles right angles. Sec page 59. J 



5. A rhombus is a quadrilateral which 
has all its sides equal, but its angles are 
not right angles. 




6. A trapezium is a quadrilateral which has 
ojie pair of parallel sides. 



PARALLELOGRAMS 57 



Theorem 20. [Euclid I. 33] 

The straight lines ivhich join the extremities of two equal and 
parallel straight lines towards the same parts are themselves 
equal and parallel. 




Let AB and CD be equal and parallel straight lines ; and 
let them be joined towards the same parts by the straight 
lines AC and BD. 

It is required to prove that AC and BD are equal and parallel. 
Join BC. 

Proof. Then because AB and CD are parallel, and BC 
meets them, 

.-. the ZABC = the alternate Z DCB. 

Now in the A ABC, DCB, 
f AB = DC, 

because { BC is common to both ; 

I and the Z ABC = the Z DCB ; Proved. 

.'. the triangles are equal in all respects ; 

so that AC = DB, (i) 

and the Z ACB = Z DBC. 

But these are alternate angles ; 

.•. AC and BD are parallel (ii) 

That is, AC and BD are both equal and parallel. 

Q.E.D. 



58 gf:ometiiy 

Theorem 21. [Euclid I. 34] 

The opposite sides and angles of a parallelogram are equal to 
one another, and each diagonal bisects the parallelogram. 
A B 




D> C 

Let ABCD be a parallelogram, of which BD is a diagonal. 
It is required to prove that 

(i) AS = CD, and AD = CB, 
(ii) the I BAD = the Z DCB, 
(iii) the Z ADC = the Z CBA, 
(iv) the AABD = the A CDB in area. 
Proof. Because AB and DC are parallel, and BD meets 
them, 

.-. the Z ABD = the alternate Z CDB. 

Because AD and BC are jiarallel, and BD meets them, 
.-. the Z .4/)i} = the alternate Z CBD. 
Hence in the A ABD, CDB, 
[the Z ABD = the Z CDB, 
because the Z A DB = the Z CBD, Proved. 

[and BD is common to both ; 
.'. tlie triangles are equal in all respects ; Thcor. 17. 

so that AB = CD, and AD = CB ; (i) 

and the Z BAD = the Z DCB ; (ii) 

and the A ABD = the A CDB in area (iv) 

And because the Z i4D^ = the Z CBD, Proved. 

and the Z CDB = the Z ABD, 
.-. the whole Z .4DC = the whole Z CBi4. . (iii) 

Q.E.D. 



PARALLELS AKD PARALLELOGRAMS 59 

CoRoi.LAUY 1. // one angle of a parallelo(jram is a right 
angle, all its angles are right angles. 
In other words: 

All the angles of a rectangle are right angles. 

For the sum of two consecutive A = 2 rt. ^*; (Theor. 14.) 
•*. , if one of these is a rt. angle, the other must be a rt. angle. 
And the opposite angles of the par"* are equal; 
.'. all the angles are right angles. 

Corollary 2. All the sides of a square are equal; and all 
its angles are right angles. 

Corollary 3. The diagonals of a 'parallelogram bisect one 
another. D C 

Let the diagonals AC, BD oi the par*^' 
A BCD intersect at 0. # ^^-^O 

To prove AG = OC, and BO = OD. 




In the A AOB, COD, A B 

f the Z OAB = the alt. Z OCD, 
because I the Z AOB = vert. opp. Z COD, 
I and AB = the opp. side CD; 
.: OA = OC; and OB = OD. Theor. 17. 



EXERCISES 

1. // the opposite sides of a quadrilateral are equal, the figure is a 
parallelogram. 

2. If the opposite angles of a quadrilateral are equal, the figure is a 
parallelogram. 

3. If the diagonals of a quadrilateral bisect each other, the figure is a 
parallelogram. 

4. The diagonals of a rhombus bisect one another at right angles. 

5. If the diagonals of a parallelogram are equal, all its angles 
are right angles. 

6. In a parallelogram which is not rectangular the diagonals are 
unequal. 



60 GEOMETRY 

EXERCISES ON PARALLELS AND PARALLELOGRAMS 

(Symmetry and Superposition) 

1. Shew that by folding a rhombus about one of its diagonals 
the triangles on opposite sides of the crease may be made to coincide. 

That is to say, prove that a rhombus is symmetrical about either 
diagonal. 

2. Prove that the diagonals of a square aro axes of symmetry. 
Name two other lines about which a square is symmetrical. 

3. The diagonals of a rectangle divide the figure into two con- 
gruent triangles : is the diagonal, therefore, an a.xis of symmetry ? 
About what two lines is a rectangle symmetrical? 

4. Is there any axis about which an oblique parallelogram is 
symmetrical ? Give reasons for your answer. 

5. In a quadrilateral ABCD, AB = AD and CB = CD; but 
the sides are not all equal. Which of the diagonals (if either) is an 
a.xis of symmetry? 

6. Prove by the method of superposition that 

(i) Tivo parallelograms are identically equal if two adjacent sides of 
one are equal to two adjacent sides of the other, each to each, aiid one 
angle of one equal to one angle of the other. 

(ii) Two rectangles are equal if two adjacent sides of one are equal 
to two adjacent sides of the other, each to each. 

7. Two quadrilaterals ABCD, EFGH have the sides AB, BC, 
CD, DA equal respectively to the sides EF, FG, GH, HE, and have 
also the angle BAD equal to the angle FEII. Shew that the figures 
may be made to coincide with one another. 

(Miscellaneous Theoretical Examples) 

8. Any straight line drawn through the middle point of a diago- 
nal of a parallelogram and terminated by a pair of opposite sides, 
is bisected at that point. 

9. In a parallelogram the perpendiculars drawn from one pair of 
opposite angles to the diagonal which joins the other pair are equal. 

10. If ABCD is a parallelogram, and .V, Y respectively the 
middle points of the sides AD, BC ; shew that the figure AY'CX is 
a parallelogram. 



PARALLELS AND PARALLELOGRAJMS 01 

11. ABC and DEF are two triangles such that AB, BC are 
respectively equal to and parallel to DE, EF; shew that AC is 
equal and parallel to DF. 

12. ABC D is a quadrilateral in which A B is parallel to DC, and 
A D equal but not parallel to BC ; shew that 

(i) the Z ^ + the Z C = 180° = the Z B + the Z D; 
(ii) the diagonal AC = the diagonal BD; 

(iii) the quadrilateral is symmetrical about the straight lino join- 
ing the middle points of AB and DC. 

13. AP, BQ are straight rods of equal length, turning at equal 
rates (both cloekrwise) about two fixed pivots A and B respectively. 
If the rods start parallel but pointing in opposite senses, shew that 

(i) they will always be parallel ; 
(ii) the line joining PQ will always pass through a fi.xed point. 

{Miscellaneous N umerical and Graphical Examples) 

14. A yacht sailing due East changes her course successively by 
63", by 78°, by 119°, and by 64°, with a view to saiUng round an 
island. What further change must be made to set her once more 
on an Easterly course? 

15. If the sum of the interior angles of a rectihneal figure is 
equal to the sum of the exterior angles, how many sides has it, and 
why? 

16. Draw, using your protractor, any five-sided figure ABCDE, 
in which 

Z 5 = 110°, Z C = 115°, Z D = 93°, Z E = 152°. 
Verify by a construction with ruler and compasses that AE is 
parallel to BC, and account theoretically for this fact. 

17. A and B are two fixed points, and two straight lines AP, 
BQ, unlimited towards P and Q, are pivoted at .4 and B. AP, 
starting from the direction A B, turns about A clockwise at the uni- 
form rate of 7^° a second ; and BQ, starting simultaneously from 
the direction BA, turns about B counter-clockwise at the rate of 
3|° a second. 

(i) In how many seconds will .4 P and BQ be parallel ? 
(ii) Find graphically and by calculation the angle between AP 
and BQ twelve seconds from the start. 

(iii) At what rate does this angle decrease? 



62 GEOMETRY ^ 

THEOKEM 22 

If there are three or more 'parallel straight lines, and the inter- 
cepts made by them on any transversal are equal, then the cor- 
respondinq intercepts on any other transversal arc also equal. 



A P/ 


V 


B 




/ 


*\ 






C Q/ 


/ 


^r 







M 


/^ 


Nz 



E r7 N \ F 



Let the parallels AB, CD, EF cut off equal intercepts PQ, 
QR from the transversal PQR ; and let XY, YZ be the cor- 
responding; intercepts cut off fj-oin any other transversal 
XYZ. 

It is required to prove that XY = Y Z. 
Through X and Y let XM and YN be drawn parallel to PR. 
Proof. Since CD and EF are parallel, and XZ meets 
them, 

.•. the Z XYM = the corresponding Z YZX. 
And since XM, YN are parallel, each being parallel to PR, 
:. the Zil/AT = the corresponding /.XYZ. 
Now the figures PM, QN are parallelograms, 
/. XM = the opp. side PQ, and YN = the opp. side QR ; 
and since \)y hypothesis PQ — QR, 
'.: XM = YN. 
Then inihe A.YMF, YNZ, 
[the Z XYM = the Z YZN, 
because I the Z MXY = the Z NYZ, 

[ and XM = YN ; 
.'. the triangles are identically equal ; Theor. 17. 

/. XY = YZ. ■ Q.E.D. 



PARALLELS AND PAKALLELOGK^UIS 



63 



Corollary, hi a triangle ABC, if a set of lines Pp, Qq, 
Rr, . . . , drawn parallel to the base, divide one side AB into 
equal parts, they also divide the other side AC into equal parts. 




Note. The lengths of the parallel.s /'/^ Qq, Rr, . . . , may thus 
be expressed in terms of the base BC. 

Through p, q, and r let pi, q'2, r'S be drawn par' to AB. 

Then, by Theorem 22, these par's divide BC into four equal parts, 
of which Pp evidently contains one, Qq two, and Rr three. 

In other words, 

Pp = i- BC; Qq = I- BC; Rr = \ • BC. 

Similarly if the given par'^ divide ^4 B into n equal parts, 

Pp = 1 . BC, Qq =-• BC, Rr =-■ BC; and so on. 

71 n n 

*^* Problem 7, p. 78, should now be worked. 

DEFINITION 

If from the extremities of a straight hne AB perpendiculars 
AX, BY are drawn to a straight hne PQ of indefinite length, 
then X Y is said to be the orthogonal projection of AB on PQ. 



P X 



"7-Q 



TP^ 



^ 



64 



GEOMETRY 




EXERCISES OX PARALLELS AND PARALLELOGRAMS 

1. The straight line drawn through the middle point of a side of a 
triangle, parallel to the base, bisects the remaining side. 

[This is an important particular case of 
Theorem 22. 

In the A A BC, if Z is the middle point of 
AB, and ZF is drawn par' to BC, we have to 
prove that AY = YC. 

Draw YX par' to AB, and then prove tin- 
^ZAY, XYC congruent.] 

2. The straight line which joins the 
middle points of two sides of a triangle is 
parallel to the third side. 

[In the A ABC, if Z, Y are the middle 
points of AB, AC, we have to prove Z}' 
par' to BC. 

Produce ZY to V, making IT equal 
to Z Y, and join C V. Prove the S^ A YZ, CY V congruent.) 

3. The straight line which joins the middle points of tivo sides of a 
triangle is equal to half the third side. 

4. Sheiv that the three straight lines which join the middle points 
of the sides of a triangle, divide it into four congrtient triangles. 

5. Any straight line drawn from the vertex of a triangle to the base 
is bisected by the line which joins the middle points of the other sides. 

6. A BCD is a parallelogram, and A', }' are the middle points of 
the opposite sides AD, BC : shew that BX and DY trisect AC. 

7. // the middle points of adjacent sides of any quadrilateral are 
joined, the figure thus formed is a parallelogram. 

8. Shew that the straight lines which join the middle points of 
opposite sides of a quadrilateral, bisect one another. 

9. From, two points A and B, and from O the mid-point be- 
tween them, perpendifuilars A P, BQ, OX are drawn to a straight 
line CD. If AP, BQ measure respectively 42 cm. and 5-8 cm., 
deduce the length of OX, and verify your result by measurement. 

Shew that OX = UAP + BQ) or i(-l/^ - f^Q)' according as A 
and B are on the same side, or on opposite sides of CD. 



PARALLELS AND PARALLELOGRAMS 65 

10. When three parallel lines cut off equal intercepts from two 
transversals, shew that of the three parallel lengths between the two 
transversals the middle one is the Arithmetic Mean of the other two. 

IL The parallel sides of a trapezium are a centimetres and b cen- 
timetres in length. Prove that the line joining the middle points of the 
oblique sides is parallel to the parallel sides, and that .its length is 
^(a + b) centimetres. 

12. OX and OY are two straight Unes, and along OA' five points 
1, 2, 3, 4, 5 are marked at equal distances. Through these points 
parallels are drawn in any direction to meet OY. Measure the 
lengths of these parallels : take their average, and compare it with 
the length of the third parallel. Prove geometrically that the 3'"'* 
parallel is the mean of all five. 

State the corresponding theorem for any odd number (2 n + 1) 
of parallels so drawn. 

13. From the angular points of a parallelogram perpendiculars 
are drawn to any straight line which is outside the parallelogram : 
shew that the sum of the perpendiculars drawn from one pair of oppo- 
site angular points is equal to the sum of those drawn from the 
other pair. 

[Draw the diagonals, and from their point of intersection suppose 
a perpendicular drawn to the given straight line.] 

14. The sum of the perpendiculars drawn from any point in the 
base of an isosceles triangle to the equal sides is equal to the perpen- 
dicular drawn from either extremity of the base to the opposite side. 

[It follows that the sum of the distances of any point in the base 
of an isosceles triangle from the equal sides is constant, that is, the 
same whatever point in the base is taken.] 

How would this property be modified if the given point were taken 
in the base produced? 

15. The sum of the perpendiculars drawn from any point within 
an equilateral triangle to the three sides is equal to the perpendicu- 
lar drawn from any one of the angular points to the opposite side, 
and is therefore constant. 

16. Equal and parallel Unes have equal projections on any 
other straight hne. 

F 



66 



GEOMETRY 



DIAGONAL SCALES 

Diagonal scales form an iniportant application of Theorem 
22. We shall illustrate their construction and use by de- 
scribing a Decimal Diagonal Scale to shew Inches, Tenths 
and Hundredths. 

A straight line AB is divided (from .4) into inches, and the 
points of division marked 0, 1, 2, . . . . The primarj^ division 
OA is subdivided into tenths, these secondary divisions being 
numbered (from 0) 1, 2, 3, ... 9. We may now read on AB 
inches and tenths of an inch. 













8 










u 








6 


1 














4 


M 
















2 





























A987654321 



In order to read hundredths, ten lines are taken at any equal, 
intervals parallel to AB ; and perpendiculars are drawn 
through 0, 1, 2, ... . 

The primary (or inch) division corresponding to 0/1 on the 
tenth parallel is now subdivided into teti equal parts ; and 
diagonal lines are drawn, as in the diagram, 
joining to the^/-.s^ point of subdivision on the 10"" parallel, 
" 1 to the second " 

" 2 (o the third " " " " ; 

and so on. 

The scale is now complete, and its use is shewn in the 
following examj^le. 

liJxmnplc. To take from the scale a Icnfilli of 2-47 iiirfica. 

(i) Place one point of the dividers at 2 in .1 B, and e.xteud them 



DIAGONAL SCALES 67 

till the other point reaches 4 in the subdivided inch A. We have 
now 2-4 inches in the dividers. 

(ii) To get the remaining 7 hundredths, move the right-hand 
point up the perpendicular through 2 till it reaches the 7^^ parallel. 
Then extend the diA-iders till the left point reaches the diagonal 4 
also on the 7^^ parallel. We have now 2-47 inches in the dividers. 

REASON FOR THE ABOVE PROCESS 

The first step needs no explanation. The reason of the 
second is found in the Corollary of Theorem 22. 

Joining the point 4 to the corresponding 
point on the tenth parallel, we have a tri- 



angle 4,4,5; of which one side 4,4 is divided 
into ten equal parts by lines parallel to 4,5. 

Therefore the lengths of the parallels be- 
tween 4,4, and the diagonal 4,5 are ^q, -f^, -^q, 
... of the base, which is 1 inch. 

Hence these lengths are 01, -02, -03, ... of *3 2 i o 
1 inch. 

Thus, by means of the scale, the length of a straight 
hne may be measured to the nearest hundredth of an inch. 

Again, if one inch-division on the scale is taken to repre- 
sent 10 feet, then 247 inches on the scale will represent 24-7 
feet. And if one inch-division on the scale represents 100 
links, then 2-47 inches will represent 247 links. Thus a 
diagonal scale is of service in preparing plans of enclosures, 
buildings, or field-works, where it is necessary that every di- 
mension of the actual object must be represented by a line 
of proportional length on the plan. 

NOTE 

The subdixision of a diagonal scale need not be decimal. 

For instance we might construct a diagonal scale to read centi- 
metres, millimetres, and q^iarters of a millimetre ; in which case we 
should take /owr parallels to the hne AB. 



68 GEOMETRY 

EXERCISES ON LINEAR MEASUREMENTS 

1. Draw straight lines whose lengths are 1-25 inches, 2-72 inches, 
3- 08 inches. 

2. Draw a Une 2- 68 inches long, and measure its length in centi- 
metres and the nearest millimetre. 

3. Draw a hne 5-7 cm. in length, and measure it in inches (to 
the nearest hundredth). Check your result by calculation, given 
that 1 cm. = 0-3937 inch. 

4. Find by measurement the equivalent of 3- 15 inches in centi- 
metres and milUmetres. Hence calculate (correct to two decimal 
places) the value of 1 cm. in inches. 

5. Draw lines 2-9 cm. and 6-2 cm. in length, and measure them 
in inches. Use each equivalent to find the value of 1 inch in centi- 
metres and milUmetres, and take the average of your results. 

6. A distance of 100 miles is represented on a map by 1 inch. 
Draw lines to represent distances of 336 miles and 408 m.iles. 

7. If 1 inch on a map represents 1 kilometre, draw lines to rep- 
resent 850 metres, 2980 metres, and 1010 metres. 

8. A plan is drawn to the scale of 1 inch to 100 links. Measure 
in centimetres and millimetres a line representing 417 Unks. 

9. Find to the nearest hundredth of an inch the length of a line 
which will represent 42-500 kilometres in a map drawn to the scale 
of 1 centimetre to 5 kilometres. 

10. The distance from London to Oxford (in a direct line) is 
55 miles. If this distance is represented on a map by 2-75 inches, 
to what scale is the map drawn ? That is, how many miles will be 
represented by 1 inch? How many kilometres by 1 centimetre? 

[1 cm. = 0-3937 inch; 1 km. = f mile, nearly.] 

11. On a map of France drawn to the scale 1 inch to 35 miles, 
the distance from Paris to Calais is represented by 4-2 inches. Find 
the distance accurately in miles, and approximately in kilometres, 
and express the scale in metric measure. [1 km. = J mile, nearly.] 

12. The distance from Exeter to Plymouth is 371 miles, and 
appears on a certain map to be 2\" ; and the distance from Lincoln 
to York is 88 km., and appears on another maj) to lx> 7 cm. C^om- 
pare the scales of these maps in miles to the inch. 

13. Draw a diagonal scale, 2 centimetres to represent 1 yard, 
shewing yards, feet, and inches. 



PRACTICAL GEOMETRY 69 



PRACTICAL GEOMETRY 



PROBLEMS 

The following problems arc to be solved with ruler and 
compasses only. No step requires the actual measurement 
of any line or angle ; that is to say, the constructions are to be 
made without using either a graduated scale of length, or a 
protractor. 

The problems are not mereh' to be studied as propositions ; 
but the construction in every case is to be actuall}' performed 
by the learner, great care being given to accuracy of drawing. 

Each problem is followed by a theoretical proof ; but the 
results of the work should always be verified by measurement, 
as a test of correct drawing. Accurate measurement is also 
required in applications of the problems. 

In the diagrams of the problems lines which are inserted 
only for purposes of proof are dotted, to distinguish them 
from lines necessary to the construction. 

For practical applications of the problems the student 
should be provided with the following instruments : 

1. A flat ruler, one edge being graduated in centimetres 
and millimetres, and the other iiTinches and tenths. 

2. Two set squar.es ; one with angles of 45°, and the 
other with angles of 60° and 30°. 

, 3. A pair of pencil compasses. 

4. A pair of dividers, preferably with screw adjustment. 

5. A semi-circular protractor. 



70 GEOMETRY 



Problem 1 

To bisect a given angle. 
B 




Let BAC be the given angle to be bisected. 
Construction. With centre A, and any radius, draw an 
arc of a circle cutting AB, AC a^i P and Q. 

With centres P and Q, and radius PQ draw two arcs cutting 
at 0. Join AO. 

Then the Z BAC is bisected by .40. 
Proof. Join PO, QO. 

In the A APO, AQO, 

!AP = AQ, being radii of a circle, 
PO = QO, " " equal circles, 
and AO in common ; 
.". the triangles are equal in all respects ; Theor. 7. 
so that the Z PAO = the Z QAO ; 
that is, the Z BAC is bisected by AO. 

XoTF. PQ lias been lakcii as the radius of \\\v arcs drawn from 
tJiP fontros /' and Q, and the inttTst'clion of thoso arcs determined 
the point O. Ami radius, however, may hv used instead of /*(.'. pro- 
vided that it is great enough to secure the intersection of the arcs. 



PROBLEMS ON LINES AND ANGLES 



71 



Problem 2 
To bisect a given straight line. 



A«- 



■♦B 



Let AB be the line to be bisected. 
Construction. With centre A, and radius AB, draw two 
arcs, one on each side of AB. 

With centre B, and radius BA, draw two arcs, one on each 
side oi AB, cutting the first arcs at P and Q. 
Join PQ, cutting AB at 0. 
Then AB is bisected at 0. 
Proof. Join AP, AQ, BP, BQ. 

In the A APQ, BPQ, 

!AP = BP, being radii of equal circles, 
AQ ^ BQ, for the same reason, 
and PQ is common ; 
.-. the Z ^PQ = the Z BPQ. Theor. 7. 

Again in the A APO, BPO, 



r 
because { 



AP = BP, and PO is common, 
I and the Z APO = the Z BPO ; 

.: AO = OB ; . Theor. 4. 

that is, AB is bisected at 0. 
Note. From the congruence of the il^ APO, BPO it follows 
that the Z AOP = the Z BOP. As these are adjacent angles, it 
follows that PQ bisects AB at right angles. 



72 



GEOMETRY 



Problem 3 

To draw a straight line perpendicular to a given straight line 
at a given point in it. 





± 


1 




\ 

\ 


r 




\ 



A P 



Q 



Let ABhe the straight hne, and X the point in it at which 
a perpendicular is to be drawn. 

Construction. With centre A" cut off from AB any two 
equal paits XP, XQ. 

With centres P and Q, and radius PQ, draw two arcs cut- 
ting at 0. 

Join XO. 

Then XO is perp. to AB. 
Proof. Join OP, OQ. 

In the A OXP, OXQ, 
I XP = XQ, by construction, 
because j OX is conmion, 

[ and PO = QO, being radii of equal circles ; 
.-. the Z OXP = the Z OXQ. Theor. 7. 

And these being adjacent angles, each is a right angle ; 
that is, XO is perp. to AB. 

Obs. If the point A' is near one entl of AB, one or olher of 
the alternative constructions on the next page should lie used. 



PROBLEMS ON LINES AND ANGLES 



73 




Problem 3. Second Method 

Construction. Take any point C 
outside AB. 

With centre C, and radius A', draw 
a circle cutting AB at D. 

Join DC, and produce it to meet 
the circumference of the circle at 0. 
Join XO. 
Then XO is perp. to AB. 
Proof. Join CX. 

Because CO = CX ; .'. the Z CXO = the Z COX 
and because CD = CX ; .-. the Z CXD = Z CDX. 
.: the whole Z DXO = the Z XOD + the Z XDO 
= i of 180° = 90°. 
/, XO is perp. to AB. 

Problem 3. Third Method 

Construction. With centre X 
and any radius, draw the arc 
CDE, cutting AB at C. 

W^ith centre C, and with the 
same radius, draw an arc, cutting 
the first arc at D. 

With centre D, and with the 
same radius, draw an arc, cut- 
ting the first arc at E. 

Bisect the Z DXE by XO. Prob. 1. 

Then XO is perp. to AB. 
Proof. Each of the A CXD, DXE is 60°; 

and the Z DXO is half of the Z DXE ; 

.: the Z CXO is 90°. 

That is, XO is perp. to AB. 




74 



GEOMETRY 



Problem 4 

To draw a straight line perpendicular to a given straight line 
from a given external point. -^ 




Let A' be the given external point from which a perpen- 
(hcular is to be drawn to AB. 

Construction. Take any point C on the side oi AB re- 
mote from A'. 

With centre A", and radius XC, draw an arc to cut AB at 
P and Q. 

With centres P and Q, and radius PA', draw arcs cutting 
at Y, on the side oi AB opposite to X. 

Join XY cutting AB at 0. 
Then XO is perp. to AB. 
Proof. Join PX, QX, PY, QY. 

In the APXY,QXY, 

iPX = QX, being radii of a circle, 
PY = QY, for the same reason^ 
and Xy is common ; 
.-. the Z PXY = the Z QXY. Theor. 7. 

Again, in the A PXO, QXO, 
f PX = QX, 

because I XO is common, 

I and the Z PXO = the Z QXO ; 

.-. the Z AOP = the Z XOQ. Theor. 4. 
And these being adjacent angles, each is a right angle, 
that is, XO is perp. to AB. 



PROBLEMS ON LINES AND ANGLES 



75 



0&.<;. When the point A' is nearly opposite one end of AB, 
one or other of the alternative constructions given below 
should be used. 



Problem 4. Second Method 

Construction. Take any point D in 
AB. Join DA', and bisect it at C. 

With centre C, and radius CX, draw 
a circle cutting AB at D and 0. 
Join XO. 

Then XO is perp. to AB. 

For, as in Problem 3, Second ^lethod, the Z XOD is a 
right angle. 




Problem 4. Third Method 

Construction. Take any two points 
D and E in AB. 

With centre D, and radius DA', draw 
an arc of a circle, on the side of AB op- 
posite to A'. 

With centre E, and radius EX, draw 
another arc cutting the former at, F. 
Join XY, cutting AB at 0. 
Then XO is perp. to AB. 

(i) Prove the A XDE, YDE 
equal in all respects by Theorem 7, 

so that the Z XDE = the Z YDE. 

(ii) Hence prove the A XDO, YDO equal in all respects 
by Theorem 4, so that the adjacent ADOX, DOY are equal. 
That is, XO is perp. to AB. 




76 GEOMETRY 



Problem 5 

At a given point in a given straight line to make an angle 
equal to a given angle. 





D B F O IQ G 



Let BAG be the given angle, and FG the given straight 
line ; and let be the point at which an angle is to be made 
equal to the Z BAC. 

Construction. With centre A, and with anj- radius, draw 
an arc cutting AB and AC at D and E. 

With centre 0, and with the same radius, draw an arc 
cutting FG at Q. 

\\"\\\i centre Q, and with radius DE, draw an arc cutting 

liie former arc at P. 

.Join OP. 

TluMi POQ is the required angle. 
Proof. Join ED, PQ. 

In the A POQ, EAD, 

10P = AE, being radii of equal circles, 
OQ = AD, for the same reason, 
PQ = ED, by construction ; 
.'. the triangles are equal in all respects ; 

so that the Z POQ = the Z EAD, Theor. 7. 



PROBLEMS ON LINES AND ANGLES 77 



Problem 

Thrnugh a given ■point to draw a straight line parallel to a 
given straight line. 

^ yO 




Let XY be the given straight hne, and the given point, 
through which a straight hne is to be drawn par' to XY. 

Construction. In A'}' take any point A, and join OA. 

Using the construction of Problem 5, at the point on 
the Hne AO make the Z AOP equal to the Z OA Y and alter- 
nate to it. 

Then OP is parallel to XY. 

Proof. Because AO, meeting the straight lines OP, XY, 
makes the alternate A POA , OA Y equal ; 
.-. OPispar^toXy. 



* * 
* 



The constructions of Problems 3, 4, and 6 are not usually 
followed in practical applications. Parallels and perpendicu- 
lars may be more quickly drawn by the aid of set squares. {See 
Lessons in Experimental Geometry, pp. 36, 42.) 



78 



GEOMETRY 



Problem 7 
To divide a given straight line into any numJ>cr nf equal parts. 




Let AB be the given straight hne, and suppose it is required 
to divide it into five equal parts. 

Construction. From A draw AC, a straight hnc of un- 
hmited length, making any angle with AB. 

From AC mark oH five equal parts of any length, AP, PQ, 
QR, RS, ST. 

Join TB ; and through P, Q, R, S draw i)ai'^ to TB, meet- 
ing AB in p, q, r, s. 

Then since the par'* Pp, Qq, Rr, Ss, TB cut off five equal 
parts from AT, they also cut off five equal parts from AB. 
(Theorem 22.) 

SECOND METHOD 

From A draw AC at any angle with 
AB, and on it mark o^ four equal parts 
AP, PQ, QR, RS, of any length. 

From B draw BD ))ar' to AC, and on 
it mark off BS', S'R', R'Q', Q'P', each 
equal to the parts marked on AC. 

Join PP', QQ', RR', SS' meeting AB 
in p, q, r, s. Then AB is divided into 
five equal parts at these points. 

[Prove by Theorems 20 and 22.] 




PROBLEMS ON LINES AND ANGLES 79 

EXERCISES ON LINES AND ANGLES 
{Graphical Exercises) 

1. Construct (with ruler and compasses only) an angle of 60°. 
By repeated bisection divide this angle into four equal parts. 

2. By means of Exercise 1, trisect a right angle; that is, divide it 
into three equal parts. 

Bisect each part, and hence shew how to trisect an angle of 45°. 
[No construction is known for exactly trisecting any angle.] 

3. Draw a line 6-7 em. long, and divide it into fire equal parts. 
Measure one of the parts in inches (to the nearest hundredth), and 
verify your work by calculation. [1 cm. = 0-3937 inch.] 

4. From a straight line 3- 72" long, cut ofif one seventh. Measure 
the part in centimetres and the nearest millimetre, and verify your 
work by calculation. 

5. At a point A' in a straight line AB draw XP perpendicular 
to AB, making XP 1-8" in length. PYom P draw an oblique PQ, 
3-0" long, to meet AB in Q. Measure XQ. 

(Problems. State your construction, and give a theoretical proof) 

6. In a straight line AT find a point whichis equidistantfrom 
two given points A and B. 

When is this impossible? 

7. In a straight line X Y find a point which is equidistant from 
two intersecting Unes AB, AC. 

When is this impossible? 

8. From a given point P draw a straight line PQ, making vnih 
a given straight Une AB a,ji angle of given magnitude. 

9. From two given points P and Q on the same side of a straight 
line A B, draw two fines Avhich meet in A B and make equal angles 
with it. 

[Construction. From P draw PH perp. to AB, and produce PH 
to P', making HP' equal to PH. Join P'Q cutting ^4 fi at A'. Join 
PK. Prove that PK, QK are the required lines.] 

10. Through a given point P draw a straight Une such that the 
perpendiculars drawn to it from two points A and B may be equal. 

Is this always possible? 



80 GEOMETRY 

THE CONSTRUCTION OF TRIANGLES 
Problem 8 
To draw a triangle having given the lengths of the three sides. 




Let a, b, c be the lengths to which the sides of the required 
triangle are to be equal. 

Construction. Draw anj^ straight line BX, and cut off 
from it a part BC equal to a. 

With centre B, and radius c, draw an arc of a circle. 

With centre C, and radius h, draw a second arc cutting 
the first at yl. 

Join AB, AC. 

Then ABC is the required triangle, for by construction the 
sides BC, CA, AB are equal to a, b, c respectively. 

Obs. The three data a, b, c may be understood in two 
ways : either as three actual lines to which the sides of the 
triangle are to be equal, or as three numbers expressing the 
lengths of those lines in terms of inches, centimetres, or some 
other linear unit. 

NoTE.s. (i) In order that the construction may be possible it is 
necessary that any two of the given sides should be together greater 
than the third side (Theorem 11); for otherwise the arcs drawn from 
the centres B and C would not cut. 

(ii) The ares which cut at A would, if eontinued, cut again on the 
other side of BC. Thus the construction gives two triangles on 
opposite! sides of a common base. 



THE CONSTRUCTION OF TRIANGLES 81 



ON THE CONSTRUCTION OF TRIANGLES 

It has been seen (page 50) that to prove two triangles 
identically equal, three parts of one must be given equal to 
the corresponding parts of the other (though any three parts 
do not necessarily serve the purpose). This amounts to 
saying that to determine the shape and size of a triangle ice 
must know three of its parts: or, in other words, 

To construct a triangle three independent data are required. 

For example, we may construct a triangle 
(i) When fivo sides {h, c) and the included angle (A) are 
given. 

The method of eonstruetion in this case is obvious. 

(ii) When two angles (A, B) and one side (a) are given. 

Here, since A and B are given, we at once know C ; 
f or A + B + C = 180°. 
Hence we have only to draw the base equal 
to o, and at its ends make angles equal to ^^-r^ 
B and C ; for we know that the remaining '^^ ' ^ 
angle must necessarily be equal to A. 



A 



(iii) If the three angles A, B, C are given (and no side), the 
problem is indeterminate, that is, the number of solutions is 
unlimited. 

For if at the ends of any base we make angles equal to 
B and C, the third angle is equal-to A. 

This construction is indeterminate, because the three data 
are not independent, the third following necessarily from the 
other two. 



82 GEOMETRY 



Problem 9 

To confitrnd a triangle having give?! two sides and an angle 
opposite to one of them. 




C, X 



Let b, c be the given sides and B the given angle. 

Construction. Take any straight line BX, and at B make 
the Z XBY equal to the given Z B. 
From BY cut off BA equal to c. 
With centre A, and radius b, draw an arc of a circle. 

If this arc cuts BX in two points Ci and d, both on the 
same side of B, both of the AABCi, ABd satisfy the given 
conditions. 

This double solution is known as the Ambiguous Case, and 
will occur when h is less than c but greater than the perp. 
from .4 on BX. 

EXERCISE 

Draw figures to illustrato the nature and number of solutions in 
the following eases : 

(i) When h is greater than c. 

(ii) When h is equal to r. 

(iii) When h is equal to the perpendicular from .1 on BX. 

(iv) When b is less than this perj)endicalar. 



THE CONSTRUCTION OF TRIANGLES 



83 



Problem 10 

To construct a right-angled triangle having given the hypot- 
enuse and otie side. 




Let AB be the hypotenuse and P the given side. 

Construction. Bisect AB at ; and with centre 0, and 
radius OA, draw a semicircle. 

With centre A, and radius P, draw an arc to cut the semi- 
circle at C. 

Join AC, BC. 
Then ABC is the required triangle. 
Proof. Join OC. 

Because OA = OC; 
:. the Z OCA = the Z OAC. 
And because OB = OC; 
.-. the Z OCB = the Z OBC. 

:. the whole Z ^CB = the Z OAC + the Z OBC 

= i of 180° Theor. 16. 

= 90°. 



84 GEOMETRY 

ON THE CONSTRUCTION OF TRIANGLES 

(Graphical Exercises) 

1. Draw a triangle whose sides are 7-5 em., 6-2 cm., and 5-3 cm. 
Draw and measure the perpendiculars dropped on these sides 

from the opposite vertices. 

2. Draw a triangle, given a = 3- 00", b - 2- 50", c - 2-75". 
Bisect the angle A by a line which meets the base at A'. Meas- 
ure BX and XC (to the nearest hundredth of an inch) ; and hence 

R V 
calculate the value of — ^ to two places of decimals. Compare your 
C -\ 

result with the value of c/b. 

3. Two sides of a triangidar field are 315 j'ards and 260 yards, 
and the included angle is known to be 39°. Draw a plan (1 inch to 
100 yards) and find by measurement the length of the remaining 
side of the field. 

4. ABC is a triangular plot of ground, of which the base BC is 
75 metres, and the angles at B and C are 47° and 68° respectivelj\ 
Draw a plan (scale 1 cm. to 10 metres). Write down without meas- 
urement the size of the angle A ; and by measuring the plan, obtain 
the approximate lengths of the other sides of the field ; also the 
perpendicular drawn from A to BC. 

5. A yacht on leaving harbour steers N.E. sailing 9 knots an 
hour. After 20 minutes she goes about, steering N.W. for 35 minutes 
and maldng the same average speed as before. How far is she now 
from the harbour, and what course (appro.ximately) must she set 
for the run home? Obtain your results from a chart of the whole 
course, scale 2 em. to 1 knot. 

6. Draw a right-angled triangle, given that the hypotenuse 
c = 10-6 cm. and one side a = 5-6 cm. Measure the third side b; 
and find the value of Vo'^-o-'. Compare the two results. 

7. Construct a triangle, having given the following parts: 
B = 34°, b =5-5 cm., c =8-5 cm. Shew that there are two solu- 
tions. Measure the two values of n, and also of C, and shew that 
the latter are supplementary. 

8. In a triangle ABC, the angle A = 50°, and ft = 6-5 cm. 
Illustrate by figures the cases which arise in constructing the triangle, 
when (i) a = 7 cm. (ii) u = 6 cm. (iii) a --= 5 cm. (iv) a = 4 cm. 



THE CONSTRUCTION OF TRIANGLES 85 

'9. Two straight roads, which cross at right angles at A, are 
carried over a straight canal by bridges at B and C. The distance 
between the bridges is 461 j'ards, and th? distance from the crossing 
A to the bridge B is 261 yards. Draw a plan, and by measurement 
of it ascertain the distance from .4 to C. 

(Problems. Stale your construction, and give a theoretical proof) 

10. Draw an isosceles triangle on a base of 4 cm., and having 
an altitude of 6-2 cm. Prove the two sides equal, and measure them 
to the nearest milhmetre. 

11. Draw an isosceles triangle having its vertical angle equal to 
a given angle, and the perpendicular from the vertex on the base 
e<iual to a given straight line. 

Hence draw an equilateral triangle in which the perpendicular 
from one vertex on the opposite side is 6 cm. ^Measure the length 
of a side to the nearest millimetre. 

12. Construct a triangle .4 BC in which the perpendicular from 
A on BC is 50 cm., and the sides AB, AC are 5-8 cm. and 90 cn^ 
respectively. Measure BC. 

13. Construct a triangle ABC having the angles at B and C 
equal to two given angles L and .V, and the perpendicular from .4 
on BC equal to a given line P. 

14. Construct a triangle .4 BC (without protractor) having given 
two angles B and C and the side b. 

15. On a given base construct an isosceles triangle having its 
vertical angle equal to a given angle L. 

16. Construct a right-angled triangle, ha\-ing given the length 
of the hypotenuse c, and the sum of the remaining sides a and b. 

If c =5-3 cm., and a +b =7-3 cm., find a and b graphically; 
and calculate the value of Va^ -i- b-. 

17. Construct a triangle, given the perimeter and the angles at 
the base. For example, a -\-b +T= 12 cm., B = 70°, C = 80°. 

18. Construct a triangle ABC from the following data: 

a = 6-5 em., 6 + c = 10 em., and B = 60°. 
Measure the lengths of b and c. 

19. Construct a triangle ABC from the following data: 

a = 7 cm., c — b = I cm., and B = 55°. 
Measure the lengths of b and c. 



86 



GEOMETRY 



THE CONSTRUCTION OF QUADRILATERALS 

It has been shewn that the shape and size of a triangle are 
completely determined when the lengths of its three sides are 
given. A quadrilateral, however, is not completely deter- 
mined by the lengths of its four sides. From what follows it 
will appear that jive independent data are required to con- 
struct a quadrilateral. 

Problem 11 

To construct a quadrilateral, given the lengths of the four 
aides, and one angle. 




Let a, h, c, d be tlie given lengths of the sides, and A the 
angle between the sides equal to a and d. 

Construction. Take any straight line AX, and cut off 
from i\ AB equal to a. . , 

Make the Z BA Y equal to the Z A. 
From .4 Y cut o^ AD equal to d. 
With centre D, and radius c, draw an aic of a circle. 
With centre /?, and radius h, draw another arc to cut the 
former at C. 

Join DC, BC. 

Then A BCD is the required quadrilateral; for by construc- 
tion the sides are equal 1o a. h, c, d, and the Z DAB is ocjual 
t.) the given angle. 



CONSTRUCTION OF QUADRILATERALS 87 

Problem 12 

To construct a parallelogram having given two adjacent sides 
and the included angle. 






Let P and Q be the two given sides, and -1 llie given angle. 

Construction 1. (With ruler and compasses.) Take a line 
A5 equal to P ; and at A make the Z B.4D equal to the Z A 
and make AD equal to Q. 

With centre D, and radius P, draw an arc of a circle. 

With centre B, and radius Q, draw another arc to cut the 
former at C. 

Then A BCD is the required par™. 

Proof. Join DB. 

In the A DCB, BAD, 

{DC^BA, 
because I CB = AD, 

[ and DB is common ; 
.-. the Z CDB = the Z ABD ; Theor. 7. 

and these are alternate angles, 
/. DC is pai-' to AB. 
Also DC ^ AB • 
.'. DA and BC are also equal and parallel. Theor. 20. 
.-. ABCD is a par"*. 

Construction 2. {JVith set squares.) Draw AB and AD as 
before ; then with set squares through D draw DC par^ to 
AB, and through B draw BC par' to AD. 

By construction ABCD is a pai™ having the required parts. 



88 



GEOMETRY 



Problem 13 

To construct a square on a given side. 




Let AB be the given side. 

Construction 1. {With ruler and compasses.) At A draw 
^A" perp. to AB, and cut off from it AD equal to AB. 

With B and D as centres, and with radius AB, draw two 
arcs cutting at C. 

Join BC, DC. 
Then A BCD is the required square. 

Proof. As ill Problem 12, ABCD may be shown (o be a 
pai-". And since the Z BA D is a right angle, the figure is a 
rectangle. Also, by construction all its sides arc equal. 

.'. A BCD is a square. 

Construction 2. {With set squares.) At A draw .4 A' perp. 
to AB, and cut off from it AD equal to AB. 

Through D draw DC par' to AB, and through B draw BC 
pai-* to AD meeting DC in C. 

Then, by construction, ABCD is a rectangle. [Def. 3, 
page 56.] 

Also it has the two adjacent sides AB, AD eciual. 

.'. it is a square. 



CONSTRUCTION OF QUADRILATERALS 89 

EXERCISES 
Ox THE Construction of Quadrilaterals 

1. Draw a rhombus each of whose sides is equal to a given 
straight Hue FQ, which is also to be one diagonal of the figure. 

Ascertain (without measurement) the number of degrees in each 
angle, giving a reason for your answer. 

2. Draw a square on a side of 2-5 inches. Pro^-e theoretically 
that its diagonals are equal ; and by measuring the diagonals to the 
nearest hundredth of an inch test the correctness of your d^a^ving. 

3. Construct a square on a diagonal of 30", and measure the 
length of each sidjo. Obtain the average of your results. 

4. Draw a parallelogram A BCD, having given that one side 
AB = 5-5 em., and the diagonals AC, BD are 8 cm., and 6 cm., 
respectively. Measure AD. 

5. The diagonals of a certain quadrilateral are equal (each 6-0 
cm.), and they bisect one another at an angle of 60°. Shew thatyife 
independent data are here given. 

Construct the quadrilateral. Name its species; and give a 
formal proof of your answer. Measure the perimeter. If the angle 
between the diagonals were increased to 90°, by how much per cent 
would the perimeter be increased? 

6. In a quadrilateral A BCD, 

AB = 5-6 cm., BC = 2-5 cm., CD = 40 cm., and DA = 3-3 cm. 
Shew that the shape of the quadrilateral is not settled by these data. 

Draw the quadrilateral when (i) A = 30°, (ii) A = 60°. Why 
does the construction fail when A = 100°? 

Determine gi-aphically the least value of A for which the con- 
struction fails. 

7. Shew how to construct a quadrilateral, ha^-ing given the 
lengths of the four sides and of one diagonal. What conditions mtist 
hold among the data in order that^Iie problem may be possible? 

Illustrate your method by constructing a quadrilateral ABCD, 
when 

(i) AB = 30", BC = 1-7", CD = 2-5", DA = 2-8", and the 
diagonal BD = 2-6". Measure AC. 

(ii) AB = 3-6 cm., BC = 7-7 cm., CD = 6-8 cm., DA =5-1 
cm., and the diagonal AC = 8-5 cm. Measure the angles at B 
and D. 



90 



GEOMETRY 



LOCI 

Definition. The locus of a point is the path traced out 
by it when it moves in accordance with some given law. 



Example 1. Suppose the point P to move so 
that its distance from a fixed point is constant 
(say 1 centimetre). 

Then the locus of P is evidently the circum- 
ference of a circle whose centre is and radius 
1 cm. 




Example 2. Suppose the point P 
moves at a constant distance (say 1 cm.) 
from a fixed straight line .4 B. 

Then the locus of P is one or other of 
two straight lines parallel to AB, on 
either side, and at a distance of 1 cm. 
from it. 



Thus the locus of a point, moving under some given con- 
dition, consists of the line or lines to which the point is 
thereby restricted ; provided that the condition is satisfied 
by every point on such line or lines, and by no other. 
• When we find a series of points which .satisfy the given 
law, and tln-ough which therefore the moving point must pass 
we are said to plot the locus of the point. 



LOCI 



91 



Problem 14 

To find the locus of a point P which moves so that its distances 
from two fixed points A and B are always equal to one another. 




Here the point P moves through all positions in which PA = 

PB; 
.'. one position of the moving point is at the middle point 

olAB. 

Suppose P to be amj other position of the moving point : 
that is, let PA = PB. 

Join OP. 

Then in the A POA, POB, 
PO is common, 
because OA = OB, 

and PA = PB, bj' hypothesis ; 
.-. the Z POA = the Z POB. Theor. 7. 

Hence PO is perpendicular to AB. * 

That is, every point P which is equidistant from A and B lies 
on the straight line bisecting AB at right angles. 

Likewise it may be proved that every point on the perpen- 
dicular through is equidistant from A and B. 
This line is therefore the required locus. 



92 



GEOMETRY 



Problem 15 

To find the locus of a point P which moves so that its perpen- 
dicular distances from two given straight lines AB, CD are equal 
to one another. 




Let P be any point such that the perp. PM = the perp. 
PN. 

Join P to 0, the intersection of AB, CD. 

Then in the A PMO, PNO, 
the A PMO, PXO are right angles, 
because the hypotenuse OP is common, 

and one side PM = one side PN ; 
.'. the triangles are equal in all respects ; Thcor. 18. 
so that the Z POM = the Z PON. 

Hence, if P lies within the Z BOD, it must he on the bisec- 
tor of that angle ; 

and, if P is within the Z AOD, it must be on the bisector of 
that angle. 

It follows that the required locus is the pair of lines which 
bisect the angles between AB and CD. 



LOCI 



93 




INTERSECTION OF LOCI 

The method of Loci may be used to find the position of a 
point which is subject to two conditions. For correspond- 
ing to each condition there will be a locus on which the re- 
quired point must lie. Hence all points which are common 
to these two loci, that is, all the points of intersection of the 
loci, will satisfy both the given conditions. 

Example 1. To find a point equidistant from three given points 
A, D, C which are not in the same straight line. 

(i) The locus of points equidistant 
from A and R is the straight line PQ, 
which bisects AB a,t right angles. 

(ii) Similarly, the locus of points equi- 
distant from B and C is the straight hne 
RS, which bisects BC at right angles. 

Hence the point common to PQ and 
RS must satisfy both conditions : that is 
to say, X the point of intersection of PQ 
and RS will be equidistant from A, B, and C. 

Example 2. To construct a triangle, having given the base, the 
altitude, and the length of the median which bisects the base. 

Let A B be the given base, and P and 
Q the lengths of the altitude and median 
respectively. 

Then the triangle is known if its vertex is 
known. 

(i) Draw a straight hue CD parallel to 
AB, and at a distance from it equal to P : P 

then the required vertex must lie on CD. 

(ii) Again, from the middle point of ^ 5 as centre, with radius 
equal to Q, describe a circle : 

theii the required vertex must lie on this circle. 

Hence any points which are common to CD and the circle satisfy 
both the given conditions : that is to say, if CD intersect the circle 
in E, F, each of the points of intersection might be the vertex of the 
required triangle. This supposes the length of the median Q to be 
greater than the altitude. 




94 GEOMETRY 

It may happen that the data of the problem are so related to one 
another that the resulting loci do not intersect. In this case the 
problem is impossible. 

Obs. In examples on the Intersection of Loci the student 
should make a point of investigating the relations which must 
exist among the data, in order that the problem may be 
possible ; and he must observe that if under certain relations 
two solutions are possible, and under other relations no solu- 
tion exists, there will always be some intermediate relation 
under which the two solutions combine in a single solution. 

EXAMPLES ON LOCI 

1. Find the locus of a point which moves so that its distance 
(measured radially) from the circumference of a given circle is 
constant. 

2. A point P moves along a straight line RQ ; find the position 
in which it is equidistant from two given points A and B. 

3. A and B are two fixed points within a circle : find points on 
the circumference equidistant from A and B. How many such 
points are there? 

4. A point P moves along a straight line RQ; find the position 
in which it is equidistant from two given straight lines AB and CD. 

5. A and B are two fixed points cm. apart. Find by the 
method of loci two points which are 4 cm. distant from A, and 5 cm. 
from B. 

6. AB and CD are two given straight lines. Find points 3 cm. 
distant from AB, and 4 cm. from CD. How many solutions are 
there? 

7. A straight rod of given length slides between two straight 
rulers placed at right angles to one another. 

Plot the locus of its middle point ; and shew that this locus is the 
fourth part of the circumference of a circle. [See Problem 10.) 

8. On a given bas(> as hypotenuse right-angled triangles are 
described. Find the locus of thi'ir vertices. 



EXAMPLES ON LOCI 95 

9. A is a fixed point, and the point X moves on a fixed straight 
line BC. ; ' 

Plol the locus of P, the middle point of AX; and prove the locus 
to be a straight line parallel to BC. 

10. A is a fixed point, and the point -Y moves on the circumfer- 
ence of a given circle. 

Plol the locus of P, the middle point of AX ; and prove that this 
locus is a circle. [See Ex. 3, p. 64.] 

11. AB is a. given straight line, and AX is the perpendicular 
drawn from A to any straight line jiassing through B. If BX re- 
volve about B, find the locus of the middle point of AX. 

12. Two straight Hnes OX, OY cut at right angles, and from P, 
a point within the angle XOY, perpendiculars PM, PN are drawn 
to OX, OY respectively. Plot the locus of P when 

(i) PM + PN is constant ( = 6 cm., say) ; 
(ii) PM — P N is constant ( = 3 cm., say). 
And in each case give a theoretical proof of the result you arrive at 
experimentallJ^ 

13. Two straight lines OX, OY intersect at right angles at 0; 
and from a movable point P perpendiculars PM, P N are drawn to 
OX, OY. 

Plot (without proof) the locus of P, when 
(i) PM = 2 PN; 
(ii) PM = 3 P iV. 

14. Find a point which is at a given distance from a given point 
and is equidistant from two given parallel straight lines. 

When does this problem admit of two solutions, when of one 
only, and when of none? 

15. S is a fixed point 2 inches distant from a given straight line 
MX. Find two points which are 2| inches distant from S, and also 
2| inches distant from MX. — 

16. Find a series of points equidistant from a given point <S' and 
a given straight line MX. Draw a curve freehand passing thi'ough 
all the points so found. 

17. On a given base construct a triangle of given altitude, hav- 
ing its vertex on a given straight line. 

18. Find a point equidistant from the three sides of a triangle. 



96 GEOMETRY 

19. Two straight lines OX, OY cut at right angles ; and Q and 
R points in OX and OY respectivelj'. Plot the locus of the middle 
point of QR, when 

(i) OQ + OR = constant; 
(ii) OQ — OR = constant. 

20. S and S' are two fixed points. Find a series of points P 
such that 

(i) SP + S'P = constant (say 3-5 inches) ; 
(ii) SP — S'P = constant (say 1-5 inches). 

In each case draw a curve freehand passing through all the points 
so found. 

ON THE CONCURRENCE OF STRAIGHT LINES IN A 
TRIANGLE 

I. The perpendiculars drawn to the sides of a triangle from their 
middle points are concurrent. 

Let AfiCbea A, and X, Y,Z the middle 
points of its sides. 

From Z and Y draw perps. to ^4 B, A C, 
meeting at 0. Join OX. 

It is required to prove that OX is pcrp. 
to BC. 

Join OA, OB, OC. 

Proof. Becau.se YO bisects AC at right angles, 

.'. it is the locus of points equidistant from A and C ; 
.: OA = OC. 

Again, because ZO bisects AB Rt right angles, 
.'. it is the locus of points equidistant from .1 and B ; 
.: OA = OB. 
Hence OB = OC. 

.'. is on the locus of points equidistant from B and C ; 

that is, OX is pcrp. to BC. 
Hence the perpendiculars from tlu; mid-points of the sides meet 
at O. 




CONCURRENCE OF LINES IN A TRIANGLE 



97 




II. The bisectors of the angles of a triangle are concurrent. 

Let ABC he a A. Bisect the A ABC, ^ 

BCA by straight lines which meet at 0. 

J 0171 AO. ' 

It is required to prove that AO bisects 
the Z BAC. 

From draw OP, OQ, OR perp. to the 
sides of the A. 

Proof. Because BO bisects the Z A BC, 

.'. it is the locus of points equidistant from BA and BC ; 

:. OP = OR. 

Similarly CO is the locus of points equidistant from BC and CA ; 

.-. OP = OQ. 
Hence OR = OQ. 

.". is on the locus of points equidistant from AB and AC; 
that is, OA is the bisector of the Z BAC. 
Hence the bisectors of the angles meet at 0. q.e.d. 

11a The bisectors of an interior arg^e.at one vertex of a triangle 
and of the exterior angles nt the other vertices are concurrent. 

Let ABC be a A, and let AB be pro- 
duced to D and AC be produced to E. 
Bisect the A CBD, BCE by straight lines 
which meet at 0. 

Join AO. 

It is required to prove that AO bisects 
the Z BAC. 

From draw OP, OQ, OR perp. to BC, 
AE, AD. _ 

Proof. As in Exercise II prove that 

OP = OR, 
OP = OQ, 
OR = OQ; 

and hence that the bisectors of the angles BAC, CBD, BCE, meet 

at 0. Q.E.D. 




98 



GEOMETRY 




III. The medians of a triangle are concurrent 
Let ABC he a A. 
Let BY and CZ be two of its medians, and 
let them intersect at O. 
Join AG, 
and produce it to meet BC in A'. 
It is required to shew that AX is the remain- k. 
ing median of the A- 
Through C draw CK parallel to BY ; 
produce ^.V to moct CK at K. 
Join BK. 

Proof. In the A AKC, 

because }' is the middle point of AC, and YO is parallel to CK, 
:. is the middle point of .1 K. Theor. 22. 

Again in the A ABK, 
since Z and O are the middle points of .4 B, A K, 
.'. ZO is parallel to BK, 
that is, OC is parallel to BK, 
.'. the figure BKCO is a par™. 
But the diagonals of a par™ bisect one another ; 
.'. A' is the middle point of BC. 
That is, .4 A' is a median of the A- 
Hence the three medians meel at the ])()int O. q.e.d. 

Dep'initiox. The point of intersection of the medians is called 
the centroid of the triangle. 

Corollary. The three 7nedians of a triangle cxt onr another at n 
point of trisection, the greater segment in each being towards the angular 
point. 

For in the above figure it has been proved that 

.40 = OK, 

also that OX is half of OK ; 
.-. OX is half of 0.4 : 
that is. O.V is one third of .4 X. 
Similarly OY is one third of BY, 
and OZ is one third of CZ. 

Q.E.O. 



CONCURRENCE OF LINES IN A TRIANGLE 99 




BNp 



N. 



\- 



* IV. The perpendiculars drawn from the vertices e/ c triangle to 
the opposite sides are concurrent. 
Let ABC be a A. 
From A, B, and C draw AD, BE, 
and CF perp. respectively to BC, 
CA, and AB. 

It is required to prove that AD, 
BE, and CF are concurrent. 

Through A draw C'B' parallel to 
BC. 

Through B and C draw CA' and 
A'B' parallel respectively to CA 
and AB. 

Proof. Because AC is parallel to 5C, 

and BC is parallel to AC, 
.*. ACBC is a parallelogram. 
.-. AC = BC. 
Similarly we may prove that AB' = BC. 

.-. A is the middle point of C'B'. 

Because the Z ADC = a right Z, 

and the line B'C is parallel to BC, 

.-. AD is perpendicular to B'C. Theor. 14, (1). 
Hence AD is perpendicular to B'C at its middle point. 
Similarly, BE and CF are perpendicular to CA' and A'B' at 
their middle points. 

.•. AD, BE, and CF are concurrent. Page 96, I. 

Q.E.D. 



MISCELLANEOUS PROBLEMS 

( A theoretical proof is to be given in each case. ) 

1. A is a given point, and J5C a given straight line. From A 
draw a straight line to make with BC an angle equal to a given angle. 

How many such lines can be drawn? 

2. Draw the bisector of an angle AOB, without using the ver- 
tex in your construction. 

3. P is a given point within the angle AOB. Draw through P 
a straight line terminated by OA and OB, and bisected at P. 



100 GEOMETRY 

4. OA, OB, OC are three straight lines meeting at 0. Draw a 
transversal terminated hy OA and OC, and bisected by OB. 

5. Through a given point A draw a straight line so that the part 
intercepted between two given parallels may be of given length. 

When does this problem admit of two solutions? When of only 
one? And when of none? 

6. In a triangle A BC inscribe a rhombus having one of its angles 
coinciding with the angle A. 

7. Use the properties of an equilateral triangle to trisect a given 
straight Une. 

8. In any triangle the shorter median bisects the greater side. 

{Construction of Triangles) 

9. Construct a triangle, having given 

(i) The middle points of the three sides, 
(ii) The lengths of two sides and of the median which bisects the 

third side, 
(iii) The lengths of one side and the medians which bisect the 

other two sides, 
(iv) The lengths of the three medians. 



AREAS 



101 



PART II 



ON AREAS 

Definitions 

1. The altitude (or height) of a parallelogram with refer- 
ence to a given side as base, is the perpendicular distance 
between the base and the opposite side. 

2. The altitude (or height) of a triangle with reference to 
a given side as base, is the perpendicular distance of the 
opposite vertex from the base. 

Note. It is clear that parallelograms or triangles which are be- 
tween the same parallels have the same altitude. 

For let A P and DQ be the alti- 
tudes of the ^ ABC, DEF, which 
are between the same parallels BF, 
GH. 

Then the fig. APQD is evidently 
a rectangle ; 

:. AP = DQ. B P C Q E F 

3. The area of a figure is the amount of surface contained 
within its bounding lines. 

4. A square inch is the area of a 
square drawn on a side one inch in 
length. 





5. Similarly a square centimetre is the area of 
a square drawn on a side one centimetre in length. 



Sq. 
era. 



The terms square yard, square foot, square metre are to be under- 
stood in the same sense. 

6. Thus the unit of area is the area of a square on a side 
of unit length. 



102 



GEOMETRY 



Theorem 23 

Area of a rectangle. // the number of units in the length of 
a rectangle is multiplied by the number of units in its breadth 
the product gives the number of square units in the area. 
D C 



L.-j-.-.v. 
• j i I- 

T--i---:- 



Let A BCD represent a rectangle whose length AB is o 
feet, and whose breadth AD is 4 feet. 

Divide AB into 5 equal parts, and BC into 4 equal parts, 
and through the points of division draw parallels to the sides. 
The rectangle A BCD is now divided into compartment^-, 
each of which represents one square foot. 

Now there are 4 rows, each containing 5 squares, 
.*. the rectangle contains 5X4 square feet. 
Similarly, if the length = a linear units, and the breadth 
= b linear units, 

the rectangle contains ab U7iits of area. 
And if each side of a square = a linear units, 

the square contains a^ units of area. 
These statements may be thus abridged : 

the area of a rectangle = length X breadth (i), 

the area of a square = (sidey (ii). 

Q.E.D. 

CoROLLAiiiES. (i) Rectangles which have equal lengths and 
equal breadths have equal areas. 

(ii) Rectangles which have equal areas and equal lengths have 
also equal breadths. 



AREA OF A RECTANGLE 103 

NOTATION 

The rectangle ABCD is said to be contained by AB, AD ; 
for these adjacent sides fix its size and shape. 

A rectangle whose adjacent sides are A^B, AD is denoted b}' 
rect. AB, AD, or by AB, AD. 

A square drawn on the side AB h denoted by sq. on AB, or 
by AB^. 

EXERCISES 

(On Tables of Length and Area) 

1. Draw a figure to shew why 

(i) 1 sq. yard = 3- .sq. feet. 
(ii) 1 sq. foot = 12- sq. inches, 
(iii) 1 sq. cm. = 10- sq. mm. 

2. Draw a figrure to shew that the square on a straight line is 
four times the square on half the line. 

3. Use squared paper to shew that the square on 1" = 10- times 
the square on 0-1". 

4. If 1" represents 5 miles, what does an area of 6 square inches 
represent ? 

EXTENSION OF THEOREM 23 

The proof of Theorem 23 here given supposes that the length and 
breadth of the given rectangle are expressed by lohole numbers ; but 
the formula holds good when the length and breadth are fractional. 

This may be illustrated thus : 

Suppose the length and breadth are 3-2 cm. and 2-4 cm.; we 
shall shew that the area is (3-2 X 2-4) sq. cm. 

For length =3-2 cm. = 32 mm. 

breadth =2-4 cm. = 24 mm. 

/ area = (32 X 24) sq. mm. = ''^^ ^ '^'^ sq. cm. 

10- 

= (3-2 X 2-4) sq. cm. 



104 GEOMETRY 

EXERCISES 

{On the Area of a Rectangle) 

Draw on squared paper the rectangles of which the length (a) 
and breadth (6) are given below. Calculate the areas, and verify by 
the actual counting of squares. 

I. a = 2", b = 3". 2. a = 1-5", b = 4". 
3. a = 0-8", b = 3-5". 4. a = 2-5", b = 1-4". 
5. a = 2-2", b = 1-5". 6. a = 1-6", b = 2-1". 
Calculate the areas of the rectangles in which 

7. a = 18 metres, 6 = 11 metres. 8. a = 7 ft., 6 = 72 in. 
9. a = 2-5 km., 6=4 metres. 10. n = \ mile, 6 = 1 inch. 

II. The area of a rectangle is 30 sq. cm., and its length is 6 em. 
Find the breadth. Draw the rectangle on squared paper ; and 
verify your work by counting the squares. 

12. Find the length of a rectangle whose area is 3-9 sq. in., and 
breadth 1-5". Draw the rectangle on squared paper; and verify 
your work by counting the squares. 

13. (i) When you treble the length of a rectangle without alter- 
ing its breadth, how many times do you multiply the area? 

(ii) When you treble both length and breadth, how many times 
do you multiply the area? 

Draw a figure to illustrate your answers ; and state a general rule. 

14. In a plan of a rectangular garden the length and breadth 
are 3-6" and 2-5", one inch standing for 10 yards. Find the area 
of the garden. 

If the area is increased by 300 sq. j'ds., the breadth remaining the 
same, what will the new length be? And how many inches will rep- 
resent it on your plan? 

15. Find the area of a rectangular enclosure of which a plan 
(scale 1 cm. to 20 metres) measures 6-5 cm. by 4-5 cm. 

16. The area of a rectangle is 1440 sq. yds. If in a plan the 
sides of the rectangle are 3-2 cm. and 4-5 cm., on what scale is the 
plan drawn ? 

17. The area of a rectangular field is 52,000 sq. ft. On a plan 
of this, drawn to the scale of 1 " to 100 ft. , the length is 3- 25". What 
is the breadth? 



EXERCISES ON RECTANGLES 



105 



Calculate the areas of the enclosures of which plans are given 
below. All the angles are right angles, and the dimensions are 
marked in feet. 



18. I 

I 

i 

t 



-iff-- 



19. 





^.-24-.,. 












12 








-*---24 --^ 





12 



-30- 



•43 > 



Calculate the areas represented by the shaded parts of the follow- 
ing plans. The dimensions are marked in feet. 

21. 



20. 



f^ v"'>',">j'jj,y>>.n >^v'v.. •>^j - ' ^ • ,.^.^>'J JM 



22. 



•*■ 15 

Width of shaded border 
uniform 1\ ft. 




24. 




23. 



Width of shad'-d border 
uniform 4 ft. 

18 > 



+ F 



H^ 



yy^/y-y' 



ki 



-30- 



25. 




106 GEOMETRY 

Theorem 24. [Euclid I. 35] 

Parallelograms on the same base and between the same parallels 
are equal in area. 

A ED P 





Let the par"" ABCD, EBCF be on the same base BC, and 
between the same par^ BC, A F. 
It is required to prove that 

the par"" ABCD = the par"" EBCF in area. 
Proof. In the A FDC, EAB, 

DC = the opp. side AB ; Theor. 21. 
( the ext. Z FDC = the int. opp. Z EAB ; Theor. 14. 
because | the int. ZDFC = the ext. ZAEB ; 

I .-. the A FDC = the A EAB. Theor. 17. 

Now, if from the whole fig. ABCF the A FDC is taken, 
the remainder is the par™ ABCD. 

And if from the whole fig. ABCF the A EAB is taken, the 
remainder is the par" EBCF. 

.'. these remainders are equal ; 
that is, the par™ ABCD = the par™ EBCF. q.e.d. 

EXERCISE 

In the above diagram the sides AD, EF overlap. Draw da- 
grams in which (i) these sides do not overlap; (ii) the ends E and 
D coincide. 

Go through the proof with these diagrams, and ascertain if it 
applies to them without change. 




I 



AREAS 107 

The Area of a Parallelogram 

Let ABCD be a parallelogram, ^ — ^ 

and ABEF the rectangle on the 
same base AB and of the same alti- 
tude BE. Then by Theorem 24, 

area of par"' ABCD = area of rect. ABEF 
= AB X BE 
= base X altitude. 

Corollary. Since the area of a parallelogram depends 
only on its base and altitude, it follows that 

Parallelograms on equal bases and of equal altitudes are equal 
in area. 

EXERCISES 
(Numerical and Graphical) 

1. Find the area of parallelograms in which 

(i) the base = 5-5 cm., and the height = 4 cm. 
(ii) the base = 2-4", and the height = 1-5". 

2. Draw a parallelogram ABCD having given AB = 2\", AD 
= 1^", and the Z. A = 65°. Draw and measure the perpendicular 

from D on AB, and hence calculate the approximate area. Why 
approximate f 

Again calculate the area from the length of AD and the i>erpen- 
dicular on it from B. Obtain the average of the two results. 

3. Two adjacent sides of a parallelogram are 30 metres and 25 
metres, and the included angle is 50°. Draw a plan, 1 cm. repre- 
senting 5 metres ; and by measuring each altitude, make two inde- 
pendent calculations of the area. Give the average result. 

4. The area of a parallelogram ABCD is 4-2 sq. in., and the base 
A 5 is 2-8". Find the height. If AD =2", draw the parallelogram. 

5. Each side of a rhombus is 2", and its area is 3- 86 sq. in. Cal- 
culate an altitude. Hence draw the rhombus, and measure one of 
its acute angles. 



108 



GEOMETRY 



Theorem 25 

The Area of a Triangle. The area of a triangle is half the 
area of the rectangle on the same base and having the same 
altitude. 



F 
Fig. I. 




Fig. 2. 



Let ABC be a triangle, and BDEC a rectangle on the same 
base BC and with the same altitude A F. 

It is required to prove that the A ABC is half the rectangle 
BDEC. 

Proof. Since AF is perp. to BC, each of the figures DF, 
EF is a rectangle. 

Because the diagonal AB bisects the rectangle DF, 

.'. the A ABF is half the rectangle DF. 

Similarly, the A AFC is half the rectangle FE. 

.'. adding these results in Fig. 1, and taking the difference in 

Fig. 2, 

the A ABC is half the rectangle BDEC. ^^ 

Corollary. A triangle is half any 'parallelogram on the 
same base and between the same parallels. 

For tho A ABC is half the rect. G_ H D A E 

BCED. 

And the rect. BCED = any par™ 
BCIIG on the same base and between 
the same par's. 

/. the A .IfiCishalf theparm BCIIG. 




AREAS 109 

THE AREA OF A TRIANGLE 

If BC and AF respectively contain a units and p units of 
length, the rectangle BDEC contains ap units of area. 
.'. the area of the A ABC = | ap units of area. 
This result may be stated thus : 

Area of a Triangle = | base X altitude. 

EXERCISES ON THE AREA OF A TRIANGLE 

(Numerical and Graphical) 

1. Calculate the areas of the triangles in which 

(i) the base = 24 ft., the height = 15 ft. 

(ii) the base = 4-8", the height = 3-5". 

(iii) the base = 160 metres, the height = 125 metres. 

2. Draw triangles from the following data. In each case draw 
and measure the altitude with reference to a given side as base : 
hence calculate the approximate area. 

(i) a = 8-4 cm., b = 6-8 cm., c = 4-0 cm. 
(ii) 6=50 cm., c = 6-8 cm., A = 65°. 
(iii) a = 6-5 cm., B = 52°, C = 76°. 

3. A BC is a triangle right-angled at C ; shew that its area = 
h BC X CA. 

Given a = 6 cm., & = 5 cm., calculate the area. 

Draw the triangle and measure the hypotenuse c; draw and 
measure the perpendicular from C on the hypotenuse ; hence cal- 
culate the approximate area. 

Note the error in your approximate result, and express it as a per- 
centage of the true value. 

4. Repeat the whole process of the last question for a right- 
angled triangle ABC, in which a = 2-8^ and b = 4-5" ; C being the 
right angle as before. 

5. In a triangle, given 

(i) Area = 80 sq. in., base = 1 ft. 8 in. ; calculate the altitude, 
(ii) Area = 10-4 sq. em., altitude = 1-6 em. ; calculate the base. 

6. Construct a triangle ABC, having given a = 30", b = 2-8", 
c = 2-6". Draw and measure the perpendicular from A on BC ; 
hence calculate the approximate area. 



no 



GEOMETRY 




Theorem 26. [Euclid I. 37] 

Triangles on the same base and between the same parallels 
{hence, of the same altitude) are equal in area. 

Let the A ABC, GBC be on the D A E Q 

same base BC and between the same 
par^BCAG. 

It is required to prove that 

the A ABC = the A GBC in area. 

Proof. If BCED is the rectangle on the base BC, and 
between the same parallels as the given triangles, 

the AABCh half the rect. BCED ; Thcor. 25. 
also the A GBC is half the rect. BCED ; 

:. the A ABC = the A GBC. q.e.d. 

Similarly, triangles on equal bases and of equal altitudes are 
equal in area. 

Theorem 27. [Euclid I. 39] 

If two triangles are equal in area, and stand on the same base 
and on the same side of it, they are between the same parallels. 

Let the A ABC, GBC, standing on 

the same base BC, be equal in area ; 

and let AF and GH be their altitudes. 

It is required to prove that AG and 

BC are par*. 

Proof. The A ABC is half the rectangle contained by BC 
and AF ; and the A GBC is half the rectangle contained 
by BC and GH ; 

/. the rect. BC, AF = the rect. BC, GH ; 

.: AF = GH. Theor. 23, Cor. 2. 

Also A F and GH are paH ; 
hence AG and FH, that is BC, are par'. q.e.d. 




AREAS 111 

EXERCISES ON THE AREA OF A TRIANGLE 
( Theoretical) 

1. ABC is a triangle and ZF is drawn parallel to the base BC, 
cutting the other sides at X and Y. Join B Y and CX ; and shew that 

(i) the A XBC = the A YBC; 
(ii) the A BXY = A CXY; 
(iii) the A ABY = the A ACX. 
If B Y and CX cut at K, shew that 

(iv) the A j5A'X= the A CKY. 

2. Shew that a median of a triangle divides it into two parts of 
equal area. 

How would you di^-ide a triangle into three equal parts by straight 
lines drawn from its vertex? 

3. Prove that a parallelogram is divided by its diagonals into 
four triangles of equal area. 

4. A BC is a triangle whose base BC is bisected at A''. If Y is 
any point in the median AX, shew that 

the A ABY = the A ACY in area. 

5. If ABCD is a parallelogram, and BP, DQ are the perpen- 
diculars from B and D on the diagonal AC, then BP = DQ. 

Also if X is any point in AC, or AC produced, 

(i) the A ADX = the A ABX; 
(ii) the A CDX = the A CBX. 

6. The straight line joining the middle points of tivo sides of a 
triangle is parallel to the third side. (Use Theorems 26 and 27.) 

7. The straight line which joins the middle points of the oblique 
sides of a trapezium is parallel to each of the parallel sides. 

8. ABCD is a parallelogram, and-^X, Y are the middle points 
of the sides AD, BC ; if Z is any point in XY, or XY produced, 
shew that the triangle AZB is one quarter of the parallelogram 
ABCD. 

9. If ABCD is a parallelogram, and X, Y any points in DC and 
AD respectively, the triangles AXB, BYC are equal in area. 

10. If ABCD is a parallelogram, and P is any point within it, 
the sum of the triangles PAB, PCDis equal to half the parallelogram. 



112 GEOMETRY 

EXERCISES ON THE AREA OF A TRIANGLE 

(Numerical and Graphical) 

1. The sides of a triangular field are 370 yds., 200 yds., and 190 
yds. Draw a plan (scale 1" to 100 yards). Draw and measure an 
altitude ; calculate the approximate area of the field in square yards. 

2. Two sides of a triangular enclosure are 124 metres and 144 
metres respectively, and the included angle is observed to be 45°. 
Draw a plan (scale 1 cm. to 20 metres). Make any necessary meas- 
urement, and calculate the approximate area. 

3. If in a triangle ABC, the area = 6-6 sq. cm., and the base 
BC = 5-5 cm., find the altitude. Hence determine the locus of A. 

If also, BA =2-6 cm., draw the triangle; iind measure CA. 

4. In a triangle ABC, given area = 3-06 sq. in., and a - 30". 
Find the altitude, and the locus of A. Given C = 68°, construct 
the triangle ; and measure 6. 

5. In a triangle ABC, BC, BA have constant lengths 6 cm. and 
5 cm. ; BC is fixed, and BA revolves about B. Trace the changes 
in the area of the triangle as the angle B increases from 0° to 180°. 

Answer by drawing a series of triangles, increasing B by incre- 
ments of 30°. Find their areas and tabulate the results. 

{Theoretical) 

6. If two triangles have two sides of one respectively equal to 
two sides of the other, and the angles contained by those sides sup- 
plementary, shew that the triangles are equal in area. Can such 
triangles ever be identically equal? 

7. Shew how to draw on the base of a given triangle an isosceles 
triangle of equal area. 

8. If the middle points of the sides of a quadrilateral are joined 
in order, prove that the parallelogram so formed [see Ex. 7, p. G4] is 
half the quadrilateral. 

9. ABC is a triangle, and R, Q the middle points of the sides 
A B, AC; shew that if BQ and C li intersect in A', the triangle BXC 
is equal to the quadrilateral AQXR. 

10. Two triangles of equal area stand on the same base but on 
opposite sides of it : shew that the straight line joining their ver- 
tices is bisected by the base, or by the base produced. 



THE AREA OF A TRIANGLE 



113 



[The method given below may be omitted from a fii-st course. 
In any case it must be postponed till Theorem 29 has been read.] 

The Area of a Triangle. Given the three sides of a triangle, 
to calculate the area. 

Example. Find the area of a triangle whose sides measure 21 m., 
17 m., and 10 m. 

Let ABC represent the given 
triangle. 

Draw AD perp. to BC, and 
denote AD by p. 

We shall first find the length 
of BD. 

Let BD = X metres ; then DC 
= 21 — X metres. 

From the right-angled A ADB, we have by Theorem 29 

,4D2 = AB-' - BD^ = 102 _ a.2. 
And from the right-angled A A DC, 

AD^ = AC- - DC^ = 172 - (21 - x)2; 
.-. 102 - a;2 = 172 - (21 - .t)2 
or, 100 - a;2 = 289 - 441 + 42 x - a;2 ; 

whence x = Q. 

Again, AD^ = AB^ - BD"-; 

or p2 = 102 - 62 = 64 ; 

.-. p = 8. 
Now Area of triangle = | base X altitude 

= (I X 21 X 8) sq. m. = 84 sq. m. 




EXERCISES 
Find the area of the triangles, whose sides are as follows : 
1. 20 ft., 13 ft., 11 ft. 2. 15 yds., 14 yds., 13 yds. 

3. 21 m., 20 m., 13 m. 4. 30 cm., 25 cm., 11 cm. 

5. 37 ft., 30 ft., 13 ft. 6. 51 m., 37 m., 20 m. 

7. If the given sides are a, b, and c units in length, prove 



(i) X 



+ r- - b"- 
2a 



(ii) p"- = c- - 



a"- + c- 



b"- 



2a 



(iii) A = i </{a + 6 -|- c)(- a -f- b + c){a - 6 + c)(o -f- b- c) 
I 



114 



GEOMETRY 



THE AREA OF QUADRILATERALS 



Theorem 28 



To find the area of 




(i) a trapezium. 

(ii) any quadrilateral 

(i) Let A BCD be a trapezium, hav- 
ing the sides AB, CD parallel. Join 
BD, and from C and D draw perpen- 
diculars CF, DE to AB. 

Let the parallel sides A B, CD meas- ^ 
ure a and b units of length, and let the height CF contain h 
units. 

Then the area of ABCD = A ABD -\- A DBC 

= IABDE + ^DCCF 

= 2 <*^ + 2 ^'^ "" 2^" ~^ ^)- 



_ 1 



height X {the sum of the parallel 



That is, 

the area of a trapezium 
sides) . 
(ii) Let ABCD be any quadrilateral. 
Draw a diagonal AC ; and from B 
and D draw perpendiculars BX, DY to 
AC. These perpendiculars are called 
offsets. 

If AC contains d units of length, and 
BX, DY p and q units respectively, 
the area of the quad' ABCD = A ABC + A ADC 

= ^ACBX -{- ^ACDY 
= ^ dp -\- i dq -= ^ dip -\- q). 
That is to say, 

the area of a quadrilateral = \ dia^gonal X {sum of offsets). 




EXERCISES ON QUADRILATERALS 



115 



EXERCISES 

{Numerical and Graphical) 

1. Find the area of the trapezium in which the two parallel 
sides are 4-7" and 3-3", and the height 1-5". 

2. In a quadrilateral A BCD, the diagonal AC = 17 feet; and 
the offsets from it to B and D are 11 feet and 9 feet. Find the area. 

3. In a plan A BCD of a quadrilateral enclosure, the diagonal 
AC measures 8-2 cm., and the offsets from it to B and D are 3-4 cm. 
and 2-6 cm. respectively. If 1 em. in the plan represents 5 metres, 
find the area of the enclosure. 

4. Draw a quadrilateral A BCD from the 
adjoining rough plan, the dimensions being 
given in inches. 

Draw and measure the offsets to A and C 
from the diagonal BD; and hence calculate 
the area of the quadrilateral. 




5. Draw a quadrilateral A BCD from the 
details given in the adjoining plan. The 
dimensions are to be in centimetres. 

Make any necessary measurements of 

your figure, and calculate its area. 

7^7 B 

6. Draw a trapezium A BCD from the following data : AB and 
CD are the parallel sides. AB =4"; AD = BC = 2"; the Z A 
= the Z B = 60°. 

Make any necessary measurements, and calculate the area. 

7. Draw a trapezium A BCD in which AB and CD are the 
parallel sides; and AB = 9 cm., CD = 3 cm., and AD = BC = 5 
cm. ^ 

Make any necessary measurement, and calculate the area. 

8. From the formula area of quadK = 5 diag. X (sum of offsets) 
shew that, if the diagonals are at right angles, 

area = | (product of diagonals) . 

9. Given the lengths, of the diagonals of a quadiilateral, and the 
angle between them, prove that the area is the same wherever they 
intersect. 



116 



GEOMETRY 




THE AREA OF ANY RECTILINEAL FIGURE 

1st Method. A rectilineal figure 
may be divided into triangles whose 
areas can be separatel}^ calculated 
from suitable measurements. The 
sum of these areas will be the area 
of the given figure. 

Example. The measurements re- 
quired to find the area of the figure 
ABCDE are AC, AD, and the offsets 
BX, DY, EZ. 

2d Method. The area of a rectilineal figure is also found 
by taking a base-line (AD in the diagram below) and offsets 
from it. These divide the figure into right-angled triangles 
and right-angled trapeziums, whose areas may be found after 
measuring the offsets and the various sections of the base-line. 

Example. Find the area of the enclosure ABCDEF from the 
plan and measurements tabulated below. U 

Yahds. 

ylD = 5G 
VC = 12 AV = 50 

AZ =40 ZE = 18 
YB = 20 ^ F = 18 

AX = 10 XF = 15 



The measurements are made from 
A along the base-line to the points 
from which the offsets spring. 



CiT^ 


N 


y 12 

/ 2 

/ 20 


'5 jc 


>v ^ 


[y 



Here A AXF= h AX X XF 
A AYB= \AY X YB 
A DZE = \DZ X ZE 
A DVC = \DV X VC 



= iX 10X15 = 
= ^ X 18 X 20 = 
= iX 16X18 = 
= \X GX12 = 



yds. 



trap^XFEZ = \ XZ X {XF + ZE) = h X 30 X 33 = 
traprayfiCK = \YV X {YB + VC) = J X 32 X 32 = 
.'., by addition, the fig. ABCDEF = 



A 

75 sq. 
180 
144 

36 
495 
512 
1442 sq. yds. 



EXERCISES ON RECTILINEAL FIGURES 



117 



EXERCISES 

1. Calculate the areas of the figures (i) and (ii) from the plans 
and dimensions (in ems.) given below. 



(i) 




AC = 6cm., AD = 5cm. 

Lengths of offsets figured 

in diagram. 




AB = BD = DA = 6cm. 
EY = CZ = lcm. 
DX = 5-2cm. 



2. Draw full size the figures whose plans and dimensions are 
given below ; and calculate the area in each case. 



(i) 




The fig. is equilateral ; 
each side to be 2J". 



AX = ir, XY = $ 
YB = li" 



3. Find the area of the figure ABCDEF from the following 
measurements and draw a plan in which 1 em. represents 20 metres. 

The Plan. 





Metrk^. 






XjO C 






180 




80 to D 


150 




40toE 


120 


50toB 


60 to F 


50 
From A 






118 GEOMETRY 

EXERCISES ON QUADRILATERALS 

( Theoretical) 

1. A BCD is a rectangle, and PQRS the figure formed by joining 
in order the middle points of the sides. 

Prove (i) that PQRS is a rhombus ; 

(ii) that the area of PQRS is half that of A BCD. 

Hence shew that the area of a rhombus is half the product of its 
diagonals. 

Is this true of any quadrilateral whose diagonals cut at right 
angles? Illustrate your answer by a diagram. 

2. Prove that a parallelogram is bisected by any straight line 
which passes through the middle point of one of its diagonals. 

Hence shew how a parallelogram ABCD may be bisected by a 
straight line drawn 

(i) through a given point P ; 
(ii) perpendicular to the side AB; 
(iii) parallel to a given line QR. 

3. In the trapezium ABCD, AB is parallel to DC; and X is the 
middle point of BC. Through X draw PQ parallel to A D to meet 
AB and DC produced at P and Q. Then prove 

(i) trapezium ABCD = Tpar^APQD. 
(ii) trapezium ABCD = twice the A AXD. 

(Graphical) 

4. The diagonals of a quadrilateral ABCD cut at right angles, 
and measure 30" and 2-2" respectively. Find the area. 

Shew by a figure that the area is the same wherever the diagonals 
cut, so long as they are at right angles. 

5. In the parallelogram ABCD, A B = 80 cm., ^D = 3-2 cm., 
and the perpendicular distance between AB and DC =30 cm. 
Draw the parallelogram. Calculate the distance between A D and 
BC ; and check your result by measurement. 

6. One side of a parallelogram is 2-5", and its diagonals are 3-4" 
and 2-4". Construct the parallelogram; and, after making any 
necessary measurement, calculate the area. 

7. ABCD is a parallelogram on a fixed base AB and of con- 
stant area. Find the locus of the intersection of its diagonals. 



EXPERIMENTAL EXERCISES 



119 



EXERCISES LEADING TO THEOREM 29 




C 



In the adjoining diagram, ABC is a triangle 
right-angled at C ; and squares are drawTi on the 
three sides. Let us compare the area of the 
square on the hypotenuse AB Avith the sum of 
the squares on the sides AC, CB which contain 
the right angle. 



1. Draw the above diagram, making AC = 3 cm., BC = 4 cm. ; 
Then the area of the square on AC = 3-, or 9 sq. cm. 1 

and the square on BC = 4-, or 16 sq. cm. J 

.'. the sum of the squares on AC, BC = 25 sq. cm. 

Now measure AB ; hence calculate the area of the square on A B, 
and compare the result with the sum already obtained. 

2. Repeat the above process, making AC = 10", BC = 2-4". 

3. If a = 15, 6 = 8, c = 17, shew arithmetically that c- =a--\-l^. 
Now draw on squared paper a triangle ABC, whose sides a, b, 

and c are 15, 8, and 17 units of length ; and measure the angle ACB. 

4. Take any triangle ABC, right- 
angled at C; and draw squares on AC, 
CB, and on the hypotenuse AB. 

Through the mid-point of the square 
on CB {i.e. the intersection of the dia- 
gonals) draw lines parallel and perpen- 
dicular to the hypotenuse, thus dividing 
the square into four congruent quadri- 
laterals. These, together with the 
square on AC, will be found exactly to 
fit into the square on AB, inThe way 
indicated by corresponding numbers. 

These experiments point to the conclusion that : 
In am' right-angled triangle the square on the hypotenuse is 
equal to the sum of the squares on the other two sides. 

A formal proof of this theorem is given on the next page. 



A/ 4 ; i 

^--. / 4 

2 /' ^^-^ 

/ 3 • 



120 



GEOMETRY 



Theorem 29. [Euclid I. 47] 

In a right-angled triangle the square described on the hypote- 
nuse is equal to the sum of the squares described on the other 
two sides. G 




Let ABC be a right-angled A, having the angle ACB Si 
rt. Z. 

It is required to prove that the square on the hypotenuse AB = 
the sum of the squares on AC, CB. 

OnAB describe the sq. ADEB ; and on AC, CB describe 
the sqq. ACGF, CBKH. 

Through C draw CL pai-* to AD or BE. 
Join CD, FB. 
Proof. Because each of the A ACB, ACG is a rt. Z , 
.•. BC and CG are in the same st. line. 
Now the rt. Z BAD = the rt. Z FAC ; 
add to each the Z CAB : 
then the whole Z CAD = the whole Z FAB. 
Then in the A CAD, FAB, 
f CA = FA, 



because 



AD = AB, 

[ and the includ(Ml ZCAD = the included ZFAB ; 
.-. the A CAD = the A FAB. Thcor. 4. 



THEOREM OF PYTHAGORAS 121 

Now the rect. AL is double of the A CAD, being on the 
same base AD, and between the same par*^ AD, CL. 

And the sq. GA is double of the A FAB, being on the same 
base FA, and between the same par"^ FA, GB. 
:. the rect. AL = the sq. GA. 

Similarly by joining CE, AK, it can be shewn that 
the rect. BL = the sq. HB. 
:. the whole sq. AE = the sum of the sqq. GA, HB : 

that is, the square on the hypotenuse AB = the sum of the 
squares on the two sides AC, CB. 

Q.E.D. 

Ohs. This is known as the Theorem of Pythagoras. The 
result established may be stated as follows : 
AB"" = BC- + CA\ 

That is, if a and h denote the lengths of the sides containing 
the right angle ; and if c denotes the hypotenuse, 

Hence a- = c^ — h- ; and 6^ = c- — a-. 

Note 1. The following important results should be noticed. 
If CL and AB intersect in 0, it has been shewn in the course of 
the proof that 

the sq. GA= the rect. AL ; 

that is, AC"^ = the rect. contained hy AB, AO (i) 

Also the sq. HB = the rect. BL ; 

that is, BC- = the rect. contained by BA, BO (ii) 

Note 2. It can be proved by superposition that squares stand- 
ing on equal sides are equal in area. 
Also we can prove conversely, 
// two squares are equal in area they stand on equal sides. 



122 



GEOMETRY 



EXPERIMENTAL PROOFS OF PYTHAGORAS'S 
THEOREM 

I. Here ABC is the given 
rt.-angled A; and ABED is 
the square on the hypotenuse 
AB. 

By drawing lines par' to the 
sides BC, Cyl, it is easily seen 
that the sq. BD is divided 
into 4 rt.-angled ^, each 
identically equal to A BC, to- 
gether with a central square. 

Hence 
sq. on hypotenuse c = 4 rt. Z'^A 
+ the central square 

= 4-hab + {a -by 

= 2ab + a^ - 2ab + b' 

= a^ + b-. 



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II. Here A BC is the given 
rt.-angled A, and the figs. 
CF, HK are the sqq. on CB, 
CA placed side by side. 

FE is made equal to DII 
or CA ; and the two sqq. CF, 
HK are cut along the lines 
BE, ED. 

Then it will be found that 
the A DHE may be placed so 
as to fill up the space ACB; 
and the A BFE may be made 
to fill the space AKD. 

Hence the two sqq. CF, 
H K may be fitted together 
so as to form the single fig. 

ABED, which will be found to be a perfect square, namely tlio 
square on the hypotenuse .1 B. 







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THEOREM OF PYTHAGORAS 123 

EXERCISES 

{Numerical and Graphical) 

1. Draw a triangle ABC, right-angled at C, having given 

(i) a = 3 cm., 6=4 em. ; 
(ii) a = 2-5 em., b = 60 cm. ; 
(iii) a = 1-2", b = 3-5". 
In each case calculate the length of the hypotenuse c, and verify 
your result by measurement. 

2. Draw a triangle A BC, right-angled at C, having given : 

(i) c = 3-4", a = 30"; [See Problem 10] 
(ii) c = 5-3 cm., b = 4-5 cm. 
In each case calculate the remaining side, and verify j'our result 
by measurement. 

{T he following examples are to be solved by calculation ; bul in each 
case a plan should be drawn on some suitable scale, and the calculated 
result verified by measurement.) 

3. A ladder whose foot is 9 feet from the front of a house 
reaches to a window-sill 40 feet above the ground. What is the 
length of the ladder? 

4. A ship sails 33 miles due South, and then 56 miles due West. 
How far is it then from its starting point? 

5. Two ships are observed from a signal station to bear respec- 
tively N.E. 60 km. distant, and N.W. 11 km. distant. How far 
are they apart? 

6. A ladder 65 feet long reaches to a point in the face of a house 
63 feet above the ground. How far is the foot from the house? 

7. B is due East of A, but at an unknown distance. C is due 
South of B, and distant 55 metres. If AC is 73 metres, find AB. 

8. A man travels 27 miles due South ; then 24 miles due West ; 
finally 20 miles due North. How far is he from his starting point ? 

9. From A go West 25 metres, then North 60 metres, then East 
80 metres, finally South 12 metres. How far are you then from A ? 

10. A ladder 50 feet long is placed so as to reach a window 48 
feet high ; and on turning the ladder over to the other side of the 
street, it reaches a point 14 feet high. Find the breadth of the 
street. 



124 



GEOMETRY 



Theorem 30. [Euclid I. 48] 

// the square described on one side of a triangle is equal to the 
sum of the squares described on the other two sides, then the angle 
contained by these two sides is a right angle. 

A O 





Let ABC be a triangle in which 

the sq. on AB = the sum of the sqq. on BC, CA. 
It is required to prove that ACB is a right angle. 

Make EF equal to BC. 
Draw FD perp. to EF, and make FD equal to CA. 

Join ED. 
Proof, Because EF = BC, 

:. the sq. on EF = the sq. on BC. 
And because FD = CA, 
.'. the sq. on FD = the sq. on CA. 
Hence the sum of the sqq. on EF, FD = the sum of the 
sqq. on BC, CA. 

But since EFD is a rt. Z 
.'. the sum of the sqq. on EF, FD = the sq. on DE: Theor. 29. 
And, by hypothesis, the sqq. on BC, CA = the sq. on AB. 
.'. the sq. on DE = the sq. on AB. 
.: DE = AB. 
Then in the A ACB, DFE, 
because AC = DF,CB = FE, and AB = DE ; 

.'. the Z ACB = the Z DFE. Theor. 7. 

But, by construction, DFE is a right angle ; 

/. the Z ACB is a right angle. q.e.d. 



THEOREM OF PYTHAGORAS AND ITS CONVERSE 125 

EXERCISES ON THEOREMS 29, 30 
( Theoretical) 

1. Shew that the square on the diagonal of a given square is 
double of the given square. 

2. In the A ABC, AD is drawn perpendicular to the base BC. 
If the side c is greater than b, shew that c- — b- = BD^ — DC^. 

3. If from any point within a triangle ABC, perpendiculars 
OX, OY, OZ are drawn to BC, CA, AB respectively: shew that 

AZ^ + BX'- + CY^ = AY^ + CX^ + BZK 
! : 4. ABC is a triangle right-angled at A ; and the sides AB, AC 
are intersected by a straight hne PQ, and BQ, PC are joined. 
Prove that 

5Q2 + PC2 = BC^ + PQ-. 

■^ In a right-angled triangle four times the sum of the squares 
on the medians drawn from the acute angles is equal to five times 
the square on the hypotenuse. 

6. Describe a square equal to the sum of two given squares. 

7- Describe a square equal to the difference between two given 
squares. 

8. Divide a straight line into two parts so that the square on 
one part may be twice the square on the other. 

9. Divide a straight hne into two parts such that the sum of 
their squares shall be equal to a given square. 

{Numerical and Graphical) 

10. Determine which of the following triangles are right-angled : 

(i) a = 14 cm., 6 = 48 em., c = 50 cm. ; 
(ii) a = 40 cm., 6 = 10 em., c = 41 cm. ; 
(iii) a = 20 cm., 6 = 99 cm., c = 101 em. 

11. A BC is an isosceles triangle right-angled at C ; deduce from 
Theorem 29 that AB"" = 2AC'-. 

Illustrate this result graphically by drawing both diagonals of 
the square on AB, and one diagonal of the square on AC. 

If AC = BC = 2", find AB to the nearest hundredth of an inch, 
and verify your calculation by actual construction and measurement. 

12. Draw a square on a diagonal of 6 cm. Calculate, and also 
measure, the length of a side. Find the area. 



1267^ 



GEOMETRY 



Problem 16 

To draw squares whose areas shall he respectively twice, three 
times, four times, . . . , that of a given square. 

Hence find graphically approximate values of V2, Vs, V4, 

■VB,.... 



Take OX, OF at right angles 
to one another, and from them 
mark off OA , OP, each one unit 
of length. Join P^. 



Then PA- = OP- + OA- =1 + 1 = 2. 
/. PA = V2. 

From OX mark off OB equal to PA, and join PB 
then PB- = OP- + 05'^ = 1 + 2 = 3. 
.-. PB = V3. 
From OX mark off OC equal to PB, and j(iin PC 
then PC^ = 0P2 + OCj = 1 + 3 = 4. 
.-. PC = V4. 




/ 



The lengths of PA, PB, PC may now be found 6y measure- 
ment ; and by continuing the process we may find Vs, Vc, 

V7, .-... 



EXERCISES ON THEOREMS 29, 30 {Coniinued) 

13. Prove the following formula : 

Diagonal of square = side X V2. 

Henr^p find to the nearest centimetre the diagonal of a square on 
a side of 50 metres. 

Draw a plan (scale 1 cm. to 10 metres) and obtain the result as 
nearly as you can by measurement. 



THEOREM OF PYTHAGORAS 127 

14. ABC is an equilateral triangle of which each side = 2m 
units, and the perpendicular from any vertex to the opposite side = p. 

Prove that p = mVS. 

Test this result graphically, when each side = 8 cm. 

15. If in a triangle a = m- — n", b = 2vin, c = m- + Ji- ; prove 
algebraically that c- = a- + b-. 

Hence by giving various numerical value to /« and rt, find sets of 
numbers representing the sides of right-angled triangles. 

16. In a triangle ABC, AD is drawn perpendicular to BC. Let 
p denote the length of AD. 

(i) If a = 25 cm., p = 12 cm., BD = 9 cm. ; find b and c. 
(ii) If b = 41", c = 50", BD = 30"; find p and a. 
And prove that Vh^ _ p2 ^ -y/c- — p^ = a. 

17. In the triangle ABC, AD is drawn perpendicular to BC. 

Prove that 

c2 - BD^ = b'- - CD'-. 

If a = 51 cm., 6 = 20 cm., c = 37 cm. ; find BD. 
Thence find p, the length of AD, and the area of the triangle 
ABC. 

18. Find by the method of the last example the areas of the 
triangles whose sides are as follows : 

(i) a = 17", b = 10", c = 9". 

(ii) a = 25 ft., & = 17 ft., c = 12 ft. 

(iii) a = 41 cm., b = 28 cm., c = 15 cm. 

(iv) a = 40 yd., 6 = 37 yd., c = 13 yd. 

19. A straight rod PQ slides between two straight rulers OX, 
OY placed at right angles to one another. In one position of the 
rod OP =5-6 cm., and OQ =3-3 cm. If in another position OP = 
40 cm., find OQ graphically ; and test the acciiracy of your drawing 
by calculation. ~^ 

20. ABC is a triangle right-angled at C, and p is the length of 
the perpendicular from C on AB. By expressing the area of the 
triangle in two ways, shew that 

pc = nb. 

Hence deduce — = — + ti- 

p2 a- b- 



128 



GEOMETRY 



1. A rectangle A BCD is said to be 
contained by two adjacent sides AB, 
AD ; for these sides fix its size and 
shape. D C 

A rectangle whose adjacent sides are A B, AD is denoted by 
the red. AB, AD; or by AB, AD. 

Similarly a square drawn on the side AB is denoted by 
the sq. on AB, or AB^. 

Geometrical illustration of algebraic identities. 
A. Geometrical illustration of {a -{- b) k = ak + hk. 
Let ST = a units of length, 



R 



TV = b units of length, 
and PS = k units of length. 



Then Area of SP, PQ = k{a + b) 
Area of SP, PR = ka 
Area of TR, RQ = kb 

.-. A- (a + 6) = ak + bk. 



Th. 
Th. 
Th. 



v 

23. 
23. 
23. 



B. Geometrical illustratioti of 
+ 6c + bd. 

Let ST = a units of length. 
Let TV = b units of length. 
Let PL = c units of length. 
And LS = d units of length 

Then Area of SP, PQ 
Area of LP, PR 
Area of MR, RQ 
Area of SL, LM 

Area of TM, MN 
:. (a + b){c -\- d) = ac + be -h ad -\- bd. 



(o + 6) (c + d) = ac + ad 





M 







gi^ii. s~ — -.-.«.-. 


' T~~ b"\J 


= (c + d)(a + b). 


Th. 23. 


= ac 


Th. 23. 


= cb 


Th. 23. 


= da 


Th. 23. 


= db 


Th. 23. 



ILLUSTRATION OF ALGEBRAIC IDENTITIES 129 









M 



■ = a- -\- b- — 2ab. 
P.- ^' --R-.Q 



Geometrical illustration of (a -{- h)- = a- -{- b~ -\- 2 nb. 

PR Q 



The area SP, PQ = the area LP, 
PR + the area TM, MN + the area 
SL, LM + the area MR, RQ. 

Hence (a + by = a- + ¥ -\- 2ab. 



D. Geometrical illustration of {a — 6)' 
Let PQ = a units of length, 

RQ = b units of length, 
then PR = (a — 6) units of lengths. 

Hence : 
the area SL, LM = the area SP, PQ 
+ the area XT, TV - the area LP, 
PQ - the area XM, MN. 

or (a - 6)2 = a^ -\-b- - 2ab. 

E. Geometrical illustration of a- — 6 

Area SP, PQ - SL, LM 
= Gnomon P, N, T 
= the area LP, PQ + 

the area TM, MN 
= the area LP, PQ + 

the area NQ, QX 
= the area LP, PX. 

Hence a- — 6^ = (a — b){a + b). 




= (a + 6)(a - 6). 




EXERCISES 

1. Illustrate kia + b + c + d + e) = ak + bk + ck + dk + ek. 

2. Illustrate (a + 6 + c)^ = a^ + 62 + c^ + 2a6 + 2ac + 26c. 

3. Illustrate A, B and C (above) by paper-folding exercises. 



130 



GEOMETRY 



Theorem 31. [Euclid II. 12] 

In an obtuse-angled triangle, the square on the side subtending 
the obtuse angle is equal to the sum of the squares on the sides 
containing the obtuse angle together with tivice the rectangle 
contained by one of those sides and the projection of the other 
side upon it 




Let ABC be a triangle obtuse-angled at C; and let AD be 
drawn perp. to BC produced, so that CD is the projection of 
the side CA on BC. [See Def. p. 63.] 
It is required to prove that 

AB^ = BC^ + CA2 + 2BC- CD. 
Proof, Because BD is the sum of the lines BC, CD, 

:. BD^ = BC + CD'~ + 2BC ■ CD.. Page 129, C. 
To each of these equals add DA-. 
Then BD^ -\- DA- = BC'^ -\- {CD^ + DA^-) + 2BCCD. 
= AB- 
= CA^~ 

= BC'- + CA'~ -\-2BC ■ CD. 

Q.E.D. 



But BD^ + DA'' 
and CD'- + DA'- 

Hence AB'- 



for the Z D is a rt. Z . 



SQUARES AND RECTANGLES 



131 



Theorem 32. [Euclid II. 13] 

In every triangle the square on the side subtending an acute 
angle is equal to the sum of the squares on the sides containing 
that angle diminished by twice the rectangle contained by one 
of those sides and the projection of the other side upon it. 





Fig. 2. 

Let ABC be a triangle in which the Z C is acute ; and let 
AD he drawn perp. to BC, or BC produced ; so that CD is 
the projection of the side CA on BC. 
It is required to prove that 

AB'- = BC^ + CA- - 2BC ■ CD. 
Proof. Since in both figures BD is the difference of the 
lines BC, CD, 

.-. BD^ = BC' -\rCD'- -2BC CD. Page 129, D 
To each of these equals add DA-. 

Then BD^ + DA'- = BC^ + (CDt-\- DA'-) - 2 BC ■ CD (i) 

But BD'- + DA- = AB'-\ 
and CD-' -\- DA' = CA' j 

Hence AB' = BC + CA' -2BC ■ CD. 

Q.E.D. 



, for the Z D is a rt. Z 



132 



GEOMETRY 





Summary of Theorems 29, 31, and 32. 

A A A 



C(D) B 



(i) If the Z ACB is obtuse, 

AB'~ = BC' + CA^ -{-2BC ■ CD. Theor. 31. 

(ii) If the Z ACB is a right angle, 

AB^ = BC^ + CA2. Theor. 29. 

(iii) If the Z ACB is acute, 

AB-" = BC + CA- -2BC ■ CD. Theor. 32. 

Observe that in (i) or (ii), if the ZACB becomes 90°, AD 

coincides with AC, and CD (the projection of CA) vanishes ; 

hence, in this case, 2 BC ■ CD = 0. 

Thus the three results maj'' be collected in one enunciation : 

The square on a side of a triangle is greater than, equal to, or 

less than the sum of the squares on the other sides, according as 

the angle contained by those sides is obtuse, a right angle, or 

aciite ; the difference in cases of inequality being twice the 

rectangle contained by one of the two sides and the projection on 

it of the other. 

EXERCISES 

1. In a triangld ABC, a = 21 cm., h — 17 cm., r = 10 cm. B3' 
liow many square centimetres does c- fall short of a' + h"-? Henco 
or other\viso calculate the projection of AC on BC. 

2. ABC is an isosceles triangle in which AB — AC; and BE 
is drawn perpendicular to AC. Shew th:it BC- = 2ACCE. 

3. In the A A BC. shew that 

(i) if the Z. C =- G0°, then c"^ = «« + ir- - ah ; 
(ii) if the Z C = 120°, then c^ = a"" -\- h"" + ab. 



SQUARES AND RECTANGLES 133 

Theorem 33. 
In any triangle the sum of the squares on two sides is equal to 
twice the square on half the third side together with twice the 
square on the median which bisects the third side. 

A 




D C 

Let ABC he :i triangle, and .'lA' the nieeliau which bisects 
the base BC. 
It is required to prove that 

AB^ + AC2 = 2 J5X2 _^ 2 AX\ 
Draw AD perp. to BC ; and consider the case in which 
AB and AC are unequal, and AD falls within the triangle. 

Then of the A AXB, AXC, one is obtuse, and the other 
acute. Let the Z AXB be obtuse. 
Then from the A AXB, 

AB'~ = BX' + AX~ + 2 BX ■ XD. Theor. 3L 
• And from the A AXC, 

AC^ = XC^ + AX^~ - 2 XC XD. Theor. 32. 
Adding these results, and remembering that XC = BX, 
we have 

AB-' + AC = 2 BX- + 2 AX\ q.e.d. 

Note. The proof may easily be adapted to the case in which 
the perpendicular AD falls outside the triangle. 

EXERCISE 

In any triangle the difference of the squares on two sides is equal to 
twice the rectangle contained by the base and the intercept between the 
middle point of the base and the foot of the perpendicular drawn from 
the vertical angle to the base. 



134 ■ GEOMETRY 

EXEHC1SP]S ON THEOREMS 31-33 

1. AB is a straight line 8 cm. in length, and from its middle 
point as centre with radius 5 cm. a circle is drawn ; if P is any 
point on the circumference, shew that 

AP'^ + BP'- = 82 sq. cm. 

2. In a triangle ABC, the base BC is bisected at A'. If a = 17 
cm., 6 = 15 cm., and c = 8 cm., calculate the length of the median 
AX, and deduce the Z A. 

3. The base of a triangle = 10 cm., and the sum of the squares 
on the other sides = 122 sq. cm. ; find the locus of the vertex. 

4. Prove that the sum of the squares on the; sides of a parallelo- 
gram is equal to the sum of the squares on its diagonals. 

The sides of a rhombus and its shorter diagonal each measure 
3"; find the longer diagonal to within -01". 

5. In any quadrilateral the squares on the diagonals are to- 
gether equal to twice the sum of the squares on the straight lines 
joining the middle points of oi)posite sides. [See Ex. 7, p. M.] 

6. ABCD is a rectangle, and O any point within it : shew that 

OA- + OC- = OB^ + 0D\ 
If .1/^ = 60", BC = 2-5", and OA^- + OC"- = 2l| sq. in., find 
the distance of from the intersection of the diagonals. 

7. The sum of the squares on the sides of a quadrilateral is 
greater than the sum of the squares on its diagonals by four times 
the square on the straight line which joins the middle points of tho 
diagonals. 

8. In a triangle A B(\ tho angles at B and C are acute; if BE, 
CF are drawn i)CTj)ondicuIar to AC, AB respectively, ijrove that 

BC'- = AB • BF + AC CE. 

9. Three times the sura of tho squares on the sides of a triangle 
is equal to four times the sum of the squares on the medians. 

10. ABC is a triangl(\ and O the point of intersection of its 
medians : sln>w that 

A/i* + BC^ + C\'P = 3(0,1 2 + UB'- + OC-'). 



PROBLEMS ON AREAS 135j 

PROBLEMS ON AREAS 
Problem 17 

To describe a parallelogram equal to a given triangle, and 
having one of its angles equal to a given angle. 





Let ABC be the given triangle, and D the given ang.e. 
It is required to describe a parallelogram equal to ABC, and 
having one of its angles equal to D. 

Construction. Bisect BC at E. 

At E in CE, make the Z CEF equal to D ; 
through A draw A FG par' to BC ; 
and through C draw CG par* to EF. 
Then FECG is the required par™. 

Proof. Join AE. 

Xow the A ABE, A EC are on equal bases BE, EC, and of 
the same altitude ; 

.-. the AABE = tlie A A EC. 

/. the A ABC is double, of the A A EC. 

But FECG is a pai-™ by construction ; 

and it is double of the A A EC, 

being on the same base EC, and between the same par'** 

EO and AG. 

.: the par'^ FECG = the A ABC ; 
and one of its angles, namely CEF, = the given Z D. 




13G GEOMETRY 

EXERCISES 

(Graphical) 

1. Draw a square on a side of 5 cm., and make a parallelogram 
of equal area on the same base, and having an angle of 45°. 

Find (i) by calculation, (ii) by measurement the length of an 
oblique side of the parallelogram. 

2. Draw any parallelogram A BCD in which AB — 2\" and 
AD — 2" \ and on the base AB draw a rhombus of equal area. 

Definition. In a parallelogram 
A BCD, if through any point K in the 
diagonal AC parallels EF, HG are 
drawn to the sides, then the figures 
EH, GF are called parallelograms 
about AC, and the figures EG, HF 
are said to be their complements. 

3. In the diagram of the preceding definition shew hij Theorem 21 
that the complements EG, HF are equal in area. 

Hence, given a parallelogram EG, and a straight line IIK, de- 
duce a construction for drawing on H K as one side a i^arallelogram 
equal and equiangular to the parallelogram EG. 

4. Construct a rectangle equal in area to a given rectangle 
CDEF, and having one side equal to a given line AB. 

U AB =6 cm., CD = 8 cm., CF = 3 cm., find by measurement 
the remaining side of the constructed rectangle. 

5. Given a parallelogram A BCD, in which AB = 2-4". AD = 
1-8", and the Z A = 5.')°. Construct a i)arallelogram of equal 
area and equiangular with A BCD, the greater side measuring 2-7". 
Measure the shorter side. 

Repeat the process, giving to A any other value, and compare 
your results. What conclusion do you draw? 

G. Draw a r(<ctangle on a side of 5 cm. equal in area to an 
equilateral triangle on a side of G cm. 

Measure the remaining side of tlie rectangle, and calculate its 
approximate! area. 



PROBLEMS ON ARKAS 137 

Problem 18 
To draio a triangle equal in area to a given quadrilateral. 




Let ABCD be the given quadrilateral. 
It is required to describe a triangle equal to ABCD in area. 
Construction. Join DB. 

Through C draw CX par' to DB, meeting AB produced 
in X. 

Join DX. 

Then DAX is the required triangle. 

Proof. Now the A XDB, CDB are on the same base DB 
and between the same par^ DB, CX; 

.'. the A XDB = the A CDB in area. 

To each of these equals add the A ADB; 

then the A DAX = the fig. ABCD. 

Corollary. In the same way it is always possible to 
draw a rectilineal figure equal to a given rectihneal figure, and 
having fewer sides by one than the given figure ; and thus 
step by step, any rectilineal figure may be reduced to a 
triangle of equal area. 

For example, in the adjoining dia- 
gram the five-sided fig. EDCBA is equal 
in area to the four-sided fig. EDXA. 

The fig. EDXA may now be reduced 
to an equal A DXY. 




138 GEOMETRY 



Problem 19 

To draw a parallelogram equal in area in a given rectilineal 
figure, and having an angle equal to a given, angle. 




W 



Let ABCD be the given rectil. fig., and E the given angle. 
Jt is required to draw a par"" equal to ABCD and having an 
angle equal to E. 

Construction. Join DB. 

Through C dvaw CF pai-' to DB, and meeting AB produced 

in F. 

Join DF. 

Then the ADAF = the fig. ABCD. Prob. 18. 

Draw the par™ AGHK equal to the A ADF, and having 

the ZKAG equal to the ZE. Prob. 17. 

Then the par"" K(! = the A ADF 

= ihohir. ABCD; 
and it has the ZKAG equal to the ZE. 

Note. If tho given reftilinoal figure has mor<> than four sides, 
it must first be redueed, step by stc^p, until it is replaeed by an 
equivalent triangle. 



PROBLEMS UN AREAS 



139 



Problem 20 

To draw a square equal in area to a given rectangle. 




B E 



X 

Let A BCD be the given rectangle. 

Construction. Produce AB to E, making BE equal to 
BC. On AE draw a semi-circle ; and produce CB to meet 
the circumference at F. 

Then 5/^ is a side of the required square. 

Proof. Let A^ be the mid-point of AE, and r the radius of 
the semi-circle. Join XF. 
Then the rect. AC = AB BE 

=^ (;• + XB){r - XB) 

- r~ - XS- (p. 129, E) 

= FB'-, from the rt. angled A FBX. 

Corollary. To describe a square equal in area to any given 
rectilineal figure. 

Reduce the given figure to a triangle of equal area. Prob. 18. 
Draw a rectangle equivalent to this triangle. Prob. 17. 
Apply to the rectangle the construction given above. 



140 GEOMETRY 

EXERCISES 

{Reduction of a Reclilineal Figure to an Equivalent Triangle) 

1. Draw a quadrilateral ABCD from the following data: 
AB = BC = 5-5 cm. ; CD = DA = 4-5 em. ; the Z A = 75'. 
Reduce the quadrilateral to a triangle of equal area. Measure 

the base and altitude of the triangle, and hence calculate the ap- 
proximate area of the given figure. 

2. Draw a quadrilateral A iSCD having given : 

AB = 2-8", BC = 3-2", CD = 3-3", DA = 3-6", and the diagonal 
BD = 30". 
Construct an equivalent triangle ; and hence find the approxi- 
mate area of the quadrilateral. 

3. On a base AB, 4 cm. in length, describe an equilateral pen- 
tagon (5 sides), having each of the angles at A and B 108°. 

Reduce the figure to a triangle of equal area; and by measur- 
ing its base and altitude, calculate the approximate area of the 
pentagon. 

4. A quadrilateral field A BCD has the following measurements : 
AB =450 metres, BC = 380 m., CD = 330 m., AD = 390 m., and 
the diagonal AC = 660 m. 

Draw a plan (scale 1 cm. to 50 metres). Reduce your plan to an 
equivalent triangle, and measure its base and altitude. Hence 
estimate the area of the field. 

{Problems. State your construction, and give a theorctic(d proof.) 

5. On the base of a given triangle construct a second triangle 
eqiial in area to the first and hstving its vertex in a given lino. 

y 6. Iloduee a triangle A BC to a triangle of equal area having 
its base BD oi given length; (D lies in BC, or BC produced.) 

V 7. Construct a triangle equal in area to a given triangle, and 
having a given .altitude. 

8. ABC is a given triangle, and A' a given point. Draw a 
triangle equal in area to ABC, ha\dng its vertex at A', and its base 
in the sam(> straight lino as BC. 




PROBLEMS ON AREAS 141 

9. Construct a triangle equal in area to the quadrilateral 
ABCD, having its vertex at a given point X in DC, and its base in 
the same straight Hne as AB. 

10. Construct a triangle equal in area to a quadrilateral ABCD 
and having two of its sides equal respectively to the diagonals of the 
quadrilateral. 

11. Shew how a triangle may be divided into n equal parts by 
straight lines drawn through one of its angular points. 

12. Bisect a triangle by a straight line drawn through a given point 
in one of its sides. 

[Let ABC be the given A, and P the 
given point in the side AB. 

Bisect AB at Z ; and join CZ, CP. 
Through Z draw ZQ parallel to CP. 
Join PQ. 
Then PQ bisects the A.] 

13. Trisect a triangle by straight lines drawn from a given point 
in one of its sides. 

[Let ABC be the given A, and A' the 
given point in the side BC. 

Trisect BC at the points P, Q. Prob. 7. 

Join AX, and through P and Q draw PH 
and Q K parallel to .LY. 

Join XH, XK. 

These straight lines trisect the A ; as 
may be shewn by joining AP, AQ.] 

It— Cut off from a given triangle a fourth, fifth, sixth, or any 
part required by a straight line drawn from a given point in one of 
its sides. — -- 

15. Bisect a qundrilntcral by a straight line drawn through an 
angular point. 

[Reduce the quadrilateral to a triangle of equal area, and join 
the vertex to the middle point of the base.] 

16. Cut off from a given quadrilateral a third, a fourth, a fifth, 
or anj' part required, by a straight line drawn through a given angu- 
lar point. 




142 GEOMETRY 



MISCELLANEOUS EXERCISES 

1. AB and AC are unequal sides of a triangle ABC ; AX is the 
median through A, AP bisects the angle BAC, and AD is the per- 
pendicular from A to BC. Prove that AF is intermediate in posi- 
tion and magnitude to AX and AD. 

2. In a triangle if a perpendicular is drawn from one extremity 
(jf the base to the bisector of the vertical angle, (i) it will make with 
either of the sides containing the vertical angle an angle equal to 
half the sum of the angles at the base ; (ii) it will make with the 
base an angle equal to half the difference of the angles at the base. 

3. In any triangle the angle contained by the bisector of the 
vertical angle and the perpendicular from the vertex to the base is 
equal to half the difference of the angles at the base. 

4. Construct a right-angled triangle, having given the hypote- 
nuse and the difference of the other sides. 

5. Construct a triangle, having given the base, the difference of 
the angles at the base, and (i) the difference, (ii) the sum, of the re- 
maining sides. 

6. Construct an isosceles triangle, having given the Imse and 
the sum of one of the equal sides and the perpendicular from the 
vertex to the base. 

7. Shew how to divide a given straight line so that the square 
on one part may be double the square on the other. 

8. A BCD is a parallelogram, and is any point without the 
angle BAD or its opposite vertical angle; shew that (he triangle 
OAC is equal to the sum of the triangles GAD, GAB. 

If G is within the angle BA D or its opposite vertical angle, shew 
that the triangle GAC is equal to the difference of the triangles 
GAD, GAB. 

0. Find the locus of the intersection of the medians of triangles 
described on a given base and of given area. 

10. On the base of a given triangle construct a second t^iangl(^ 
(•(lual in area to the first, and having its vertex in a given straight 
line. 

11. A BC D is a parallelogram nuide of rods connected by hinges. 
If AB is fixed, find the locus of the middle point of CD. 



PART III 
THE CIRCLE 

Definitions and First Principles 

1. A circle is a plane figure contained by a line traced out 
by a point which moves so that its distance from a certain 
fixed point is always the same. 

The fixed point is called the centre, and the bounding line 
is called the circumference. 

XoTE. According to this definition the term circle strictly ap- 
plies to the figure contained by the circumference ; it is often used, 
however, for the circumference itself when no confusion is likely to 
arise. 

2. A radius of a circle is a straight line drawn from the 
centre to the circumference. It follows that all radii of a 
circle are equal. 

3. A diameter of a circle is a straight line drawn through 
the centra and terminated both ways by the circumference. 

4. A semi-circle is the figure-bounded by a diameter of 
a circle and the part of the circumference cut off by the 
diameter. 

It will be proved on page 146 that a diameter di^^des a circle into 
two identically equal parts. 

5. Circles that have the same centre are said to be con- 
centric. 

143 



144 GEOMETRY 

From these definitions wc draw the following inferences : 

(i) A circle is a closed curve; so that if the circumference 
is crossed by a straight line, this line if produced will cross 
the circumference at a second point. 

(ii) The distance of a point from the centre of a circle 
is greater or less than the radius according as the point is 
without or within the circumference. 

(iii) A point is outside or inside a circle according as its 
distance from the centre is greater or less than the radius. 

(iv) Circles of equal radii are identically equal. For by 
superposition of one centre on the other the circumferences 
must coincide at every point. 

(v) Concentric circles of unequal radii cannot intersect, for 
the distance from the centre of every point on the smaller 
circle is less than the radius of the larger. 

(vi) If the circumferences of two circles have a common 
point they cannot have the same centre, unless they coincide 
altogether. 

C. An arc of a circle is any part of the circumference. 

7. A chord of a circle is a straight line, joining any two 
points on the circumference. 

Note. From these definitions it may be seen 
that a chord of a circle, which does not pass 
through the centre, divides the (arcumferenco 
into two unequal arcs; of these, the greater is 
called the major arc, and the h'ss th«' minor arc. 
Thus the major arc is greater, and the minor arc 
less than the semi-circumference. 

The major and minor arcs, into wliicli a <'ir- 
cumferenee is divided by a chord, are said to be conjugate to one 
another. 




SYMMETRY OF A CIRCLE 145 

Symmetry 

Some elementary properties of circles are easily proved by 
considerations of symmetry. For convenience the definition 
given previously is here repeated. 

Definition 1. A figure is said to be symmetrical about a 
line when, on being folded about that line, the parts of the 
figure on each side of it can be brought into coincidence. 

The straight line is called an axis of symmetry. 

That this may be possible, it is clear that the two parts of the 
figure must have the same size and shape, and must be similarly 
placed with regard to the axis. 

Definition 2. Let AB he a straight line and P a point 
outside it. 

P 



Q 

From P draw PM perp. to AB, and produce it to Q, mak- 
ing MQ equal to PM. 

Then if the figure is folded about AB, the point P may be 
made to coincide with Q, for^ the Z AAIP = the Z AAIQ 
and MP = MQ. 

The points P and Q are said to be symmetrically opposite 
with regard to the axis AB, and each point is said to be the 
image of the other in the axis. 

Note. A point and its image are equidistant from every point 
on the axis. See Prob. 14, page 91. 

L 



146 GEOMETRY 

A circle is symmetncal about any diameter. 




Let APBQ be a circle of which is the centre, and AB 
any diameter. 

It is required to prove that the circle is symmetrical about 
AB. 

Proof. Let OP and OQ be two radii making any equal 
A AOP, AOQ on opposite sides of OA. 

Then if the figure is folded about AB, OP may be made to 
fall along OQ, since the Z AOP = the Z AOQ. 

And thus P will coincide with Q, since OP = OQ. 

Thus every point in the arc APB must coincide with some 
point in the arc AQB; that is, the two parts of the circum- 
ference on each side of AB can be made to coincide. 
.'. the circle is symmetrical about the diameter AB. 

Corollary. If PQ is diawn cutting AB at M, then on 
folding the figure about A/:?, since P falls on Q, MP will 
coincide with MQ, 

.'. MP = MQ; 
and the Z OMP will coincide with the Z OMQ; 
.'. these angles, being adjacent, are rt. A ; 
.'. the points P and Q are symmetrically opposite with 
regard to AB. 

Hence, conversely, // a circle passes through a given point P, 
it also pa.s.sr.s through the symmetrically opposite point with re- 
gard to any diameter. 



EQUAL CIRCLES 



147 



SOME PROPERTIES OF EQUAL CIRCLES 

The student should prove for himself the following proper- 
ties of equal circles. A, B, D, and E may readily be proven 
by superposition, while C is a simple exercise on Theorem 7. 

A. In equal circles angles at the centre lohich stand on equal 
arcs are equal. 

A D 





B. In equal circles arcs which subtend equal angles at the 
centre are equal. 

C. In equal circles arcs ivhich are cut off by equal chords 
are equal, the major arc to the major and the minor arc to the 
minor. 

A D 

(a) Prove Z BOC = 

Z EGF, Th. 7. 

(b) Hence arc BC = 

arc EF, Th. 2. 



D. In equal circles chords which cut off equal arcs are equal. 

E. hi equal circles sectors (see Def. p. 161) which have equal 
angles are equal. 

Note. State and prove these properties for the same circle. 





148 GEOMETRY 

ON CHORDS 
""■^ Theorem 34. [Euclid III. 3] 

// a straight line drawn from the centre of a circle bisects a 
chord which does not pass through the ceyitre, it cuts the chord at 
right angles. 

Conversely, if it cuts the chord at rigid angles, it bisects it. 




Let ABC be a circle whose centre is 0; and let OD bisect 
a chord AB which does not pass through the centre. 
It is required to prove that OD is perp. to AB. 

JoinO^, OB. 
Proof. Then in the A ADO, BDO, 

[ AD = BD, by hypothesis, 
because < OD is common, 

[ and OA = OB, being radii of the circle ; 
.-. the Z ADO = the Z BDO, Theor. 7. 

and these are adjacent angles; 

.'• OD is perp. to AB. q.e.d. 

Conversely. Let OD be perp. to the chord AB. 
It is required to prove that OD bisects AB. 
Proof. In the A ODA, ODB, 

the A ODA, ODB are right angles, 

the hypotenuse OA = the hypotenuse OB, 

and OD is common ; 

.'. DA = DB; Theor. 18. 

that is, OD bisects AB at D. q.e.d. 



because 



CHORD PROPERTIES l49 

Corollary 1. The straight line which bisects a chord at 
right angles passes through the centre. 

Corollary 2. A straight line cannot meet a circle at more 
than two points. 

For suppose a st. line meets a q 

circle whose centre is at the points 
A and B. 

Draw OC per p. to AB. 

Then ^C = C5. A ^; b u 

Now if the circle were to cut AB in a third point D, AC 
would also be equal to CD, which is impossible. 

Corollary 3. A chord of a circle lies wholly mthin it. 

EXERCISES 
(Numerical and Graphical) 

1. In the figure of Theorem 34:, ii AB = S cm., and OD = Z 
cm., find OB. Draw the figure, and verify your result by measure- 
ment, o v3» - 4."i - 

2. Calculate the length of a chord which stands at a distance 
G" from the centre of a circle whose radius is 13". 

3. In a circle of 1" radius draw two chords 1-6" and 1-2" in 
length. Calculate and measure the distance of each from the centre. 

4. Draw a circle whdse diameter is 8-0 cm. and place in it a 
chord 6-0 cm. in length. Calculate to the nearest millimetre the 
distance of the chord from the centre; and verify ypur result by 
measurement. 

5. Find the distance from the centre to a chord 5 ft. 10 in. in 
length in a circle whose diameter is 2 yds. 2 in. Verify the result 
graphically by drawing a figure in which 1 cm. represents 10". 

6. ^fi is a chord 2-4" long in a circle whose centre is and 
whose radius is 1-3"; find the area of the triangle OAB in square 
inches. 

7. Two points P and Q are 3" apart. Draw a circle with radius 
1-7" to pass through P and Q. Calculate the distance of its centre 
from the chord PQ, and venfy by measiiroment. 



150 



GEOMETRY 



2> \ . Theorem 35 

One circle, arid only one, can pass through atiy three points 
not in the same straight line. 




Let A, B,C be three points not in the same straight Hne. 
It is required to prove that one circle, and only one, can poffs 

through A, B, and C. 

Join AB, BC. 

Let AB and BC be bisected at right angles by the Hnes 
DF, EG. 

Then since AB and BC are not in the same st. hne, DF and 
EG are not par*. 

Let DF and EG meet in 0. 

Proof. Because DF bisects AB at right angles, 
.*. evej-y point on DF is equidistant from A and B. 

Proh. 14. 

Similarly every point on EG is equidistant from B and C. 

.'. (), the only point connnon to DF and EG, is equidistant 
from A, B, and C; 

and there is no other point equidistant from A, B, and C. 

.-. a circle having its centre at O and radius OA will ]iass 
through B and C; and this is the only circle which will pass 
through the three given i)oints. q.e.d. 



CHORD PROPERTIES 151 

Corollary 1. The size and position of a circle are fully 
determined if three of its points are known; for then the posi- 
tion of the centre and length of the radius can be found. 

Corollary 2. Two circles cannot cut one another in more 
than two points without coinciding entirely ; for if they cut at 
three points they would have the same centre and radius. 

Hypothetical Construction. From Theorem 35 it ap- 
pears that we may suppose a circle to be drawn through any three 
points not in the same straight line. 

Thus, one circle passes through the vertices of any triangle. 

Definition. The circle passing through the vertices of 
a triangle is said to be circumscribed about the triangle. 
The circle, its centre, and its radius are called the circum- 
circle,the circum-centre,and the circum-radius of the triangle. 

EXERCISES OX THEOREMS 34 AND 35 

( Theoretical) 

1. The parts of a straight line intercepted .between the circum- 
ferences of two concentric circles are equal. 

2. Two circles, whose centres are at A and B, intersect at C\ 
D ; and M is the middle point of the common chord. Shew that 
A M and BM are in the same straight line. 

Hence prove that the line of centres bisects the common chord at right 
angles. 

3. AB, AC are two equal chords of a circle; shew that the 
.straight line which bisects the angle 5,4 C passes through the centre. 

4. Find the locus of the centres of all circles which pass through tiro 
given points. 

5. Describe a circle that shall pass through two given points and 
hare its centre in a given straight line. 

When is this impossible? 

6. Describe a circle of given radius to pass through two given points. 
When is this impossible? 



152 GEOMETRY 

3S *Theorem 36. [Euclid III. 9] 

// from a point within a circle more than two equal straight 
lines can he drawn to the circumference, that point is the centre 
of the circle. 




Let ABC be a circle, and a point within it from which 
more than two equal st. lines are drawn to the O**, namely 
OA,OB,OC. 

It is required to prove that is the centre of the circle ABC. 

Join AB, BC. 
Let D and E be the middle points o( AB and BC re- 
spectively. 

Join OD, OE. 

Proof. In the A ODA, ODB, 

r DA = DB, 
because I DO is common, 

[and OA = OB, by hypothesis; 
.-. the Z ODA = the Z ODB; Thcor. 7. 
..'. these angles, being adjacent, are rt. A . 

Hence DO bisects the chord AB at right angles, and there- 
fore passes through the centre. Theor. 34, Cor. 1. 

Similarly it may be shewn that EO passes through the 
centre. 

.". O, whicli is the only point common to DO and EO, must 
be the centre. q.e.d. 



CHORD PROPERTIES 153 

EXERCISES OX CHORDS 
{Numerical and Graphical) 

1. AB and BC are lines at right angles, and their lengths are 
1-6" and 3-0" respectively. Draw the circle through the points A, 
B, and C ; find the length of its radius, and verify your result by 
measurement. 

2. Draw a circle in which a chord G em. in length stands at a 
distance of 3 cm. from the centre. 

Calculate (to the nearest millimetre) the length of the radius, 
and verify your result by measurement. 

3. Draw a circle on a diameter of 8 cm., and place in it a chord 
equal to the radius. 

Calculate (to the nearest millimetre) the distance of the chord 
from the centre, and verify bj' measurement. 

4. Two circles, whose radii are respectively 26 inches and 2.5 
inches, intersect at two points which are 4 feet apart. Find the 
distance between their centres. 

Draw the figure (scale 1 cm. to 10"), and verify your result by 
measurement. 

5. Two parallel chords of a circle whose diameter is 13" are re- 
spectively 5" and 12" in length; shew that the distance between 
them is either 8-5" or 3- 5". 

6. Two parallel chords of a circle on the same side of the centre 
are 6 cm. and 8 cm. in length respectively, and the perpendicular 
distance between them is 1 cm. Calculate and measure the radius. 

(Theoretical) 

7. The line joining the middle points of two parallel chords of a 
circle passes through the centre. 

8. Fijid the locus of the middle points of parallel chords in a circle. 

9. Two intersecting chords of a circle cannot bisect each other 
unless each is a diameter. 

10. If a parallelogram can be inscribed in a circle, the point of 
intersection of its diagonals must be at the centre of the circle. 

11. Shew that rectangles are the onlj^ parallelograms that can 
be inscribed in a circle. 



154 



GEOMETRY 



^^ Theorem 37. [Euclid III. 14] 

Equal chords of a circle are equidistant from the centre. 
Conversely, chords which are equidistant from the centre are 
equal. 




Let AB, CD be chords of a circle whose centre is 0, and let 
OF, OG be perpendiculars on them from 0. 
First. Let AB = CD. 

It is required to prove that AB and CD are equidistajif from 0. 

Join 0.4, OC. 
Proof. Because OF is perp. to the chord AB, 

.: OF bisects .4/?; Theor. 34. 

.-. .4 F is half of .4 B. 
Similarly CO is half of CD. 
But, by hypothesis, AB = CD, 
.: AF = Ca. 
Now in the A OF A, OCC, 
I the A OFA, OGC are right amjles, 
because I the hypotenuse OA = the hypotenuse OC, 
1 ' :uu\ AF = CO;' 

.'. the trian}2;les are ecjual in all respects; Thcor. 18. 
so that OF = 00 ; 
that is, AB and CD ;ire e(iuidistant from O. 

Q.E.D. 



CHORD PROPERTIES 155 

Conversely. Let OF = OC. 

li is required to prove that AB = CD. 
Proof. As before it may be shewn that AF is half of AB, 
and Cti half of CD. 

Then in the A OF A, OGC, 
[ the A OF Ay, OGC are right angles, 
because the hypotenuse OA = the hypotenuse OC, 
[ and OF = 00; 

.: AF = CO; Theor. 18. 

.'. the doubles of these are equal ; 

that is, AB = CD. q.e.d. 

EXERCISES 

( Theoretical) 

1. Find the locus of the middle points of equal chords of a circle. 

2. If two chords of a circle cut one another, and make equal 
angles with the straight line which joins their point of intersection 
to the centre, they are equal. • 

3. If two equal chords of a circle intersect, shew that the seg- 
ments of the one are equal respectively to the segments of the other. 

4. In a given circle draw a chord which shall be equal to one 
given straight line (not greater than the diameter) and parallel to 
another. 

5. PQ is a fixed chord in a circle, and AB is any diameter : shew 
that the sum or difference of the perpendiculars let fall from A and 
B on PQ is the same for all positions of AB. [See Ex. 9, p. 64.] 

{Graphical) 

6. In a^circle of radius 4- 1 cm. any number of chords are drawn 
each 1-8 cm. in length. Shew that the middle points of these chords 
all lie on a circle. Calculate and measure the length of its radius, 
and draw the circle. 

7. The centres of two circles are 4" apart, their common chord 
is 2-4" in length, and the radius of the larger circle is 3-7". Give a 
construction for finding the points of intersection of the two circles, 
and find the radius of the smaller circle. 



156 GEOMETRY 

v3 6 . Theorem 38. [Euclid III. 15] 

Of any two chords of a circle, that which is nearer to the cetitre 
is greater than one more remote. 

Conversely, the greater of two chords is nearer to the centre 
than the less. 

° .A 




Let AB, CD be chords of a circle whose centre is 0, and let 
OF, OG be perpentliculars on them from 0. 

It is required to prove that 

(i) if OF is less than OG, then AB is greater than CD ; 
(ii) if AB is greater than CD, then OF is less than OG. 
Join OA, OC. 

Proof. Because OF is perp. to the chord AB, 
.: OF bisects A B; 
.-. AFmhaUoiAB. 
Similarly CG is half of CD. 

Now OA =- OC; 
.'. the sq. on OA = the sq. on OC. 

But since the Z OFA is a rt. angle, 
/. the sq. on OA = the sqq. on OF, FA. 

Siniihirly the sq. on OC = the sqq. on OG, GC. 
.: the sqq. on OF, FA = the sqq. on OG, GC. 



CHORD PROPERTIES 157 

(i) Hence if OF is given less than OG, 

the sq. on OF is less than the sq. on OG. 
.'. the sq. on FA is greater than the sq. on GO; 

:. FA is greater than GC ; 

.•. AB is greater than CD. 

(ii) But if A 5 is given greater than CD, 
that is, if FA is greater than GC; 
then the sq. on FA is greater than the sq. on GC. 
.". the sq. on OF is less than the sq. on OG; 

.-. OF is less than OG. q.e.d. 

Corollary. The greatest chord in a circle is a diameter. 

EXERCISES 
( Miscellaneous ) 

1. Through a given point within a circle draw the least possible 
chord. 

2. Draw a triangle ^fiC in which a = 3-5", b = 1-2", c = 3-7". 
Through the ends of the side a draw a circle with its centre on the 
side c. Calculate and measure the radius. 

3. Draw the circum-eircle of a triangle whose sides are 2-6", 
2-8", and 3-0". Measure its radius. 

4. AB is a fixed chord of a circle, and XY any other chord 
having its middle point Z on AB; what is the greatest, and what 
the least length that XY may have? 

Shew that XY increases, as Z approaches the middle point of 
AB. 

5. Describe the change of direction of the chord X Y (in Ex. 4) 
as Z moves from one end of AB to its middle point. 

6. What direction docs XY take when Z reaches the middle 
point of AB ? 

7. Consider the position of X F when Z gets very near to A. 
Note. For exercises on Theorems 34-38, see page 160. 



158 



GEOMETRY 



ON ANGLES IN SEGMENTS, AND ANGLES AT 

THE CENTRES AND CIRCUMFERENCES 

OF CIRCLES 



s r 



Theorem 39. [Euclid IIL 20] 



The angle at the centre of a circle is douhle of an angle at the 
circumference standing on the same arc. 





Fig. I. 



Fig. 2. 



Let ABC be a circle, of which is the centre ; and let 
BOC be the angle at the centre, and BAC an angle al the 
O*'^ standing on the same arc BC. 
It is required to prove that the Z BOC is twice the Z BAC. 

Join AO, and produce it to D. 
Proof. In the A OAB, because OB = OA, 
.'. the Z OAB = the Z OB A. 
.'. the sum of the A OAB, OBA = twice the Z OAB. 
But the ext. Z BOD = the sum of the A OAB, OBA; 
.'. the Z BOD = twice the Z OAB. 

Similarly the Z DOC = twice the Z OAC. 
.: , adding these results in Fig. 1 , and taking the difference 
in Fig. 2, it follows in each case that 

the Z BOC = twice the Z B,\C. q.e.d. 




159 



Fig. 3. Fig. 4. 

Obs. If the arc EEC, on wliicli the angles stand, is a semi- 
chcumference, as in Fig. 3, the Z BOC at the centre is a 
straight angle ; and if the arc BEC is greater than a semi- 
circumference, as in Fig. 4, the Z BOC at the centre is reficx. 
But the proof for Fig. 1 appHes without change to both these 
cases, shewing that whether the given arc is greater, than, 
equal to, or less than a semi-circumference, 

the Z BOC = twice the Z BAC, on the same arc BEC. 

DEFINITIONS 

A segment of a circle is the figure bounded 
by a chord and one of the two arcs into which 
the chord divides the circumference. 

Note. The chord of a segment is sometimes 
called its base. 

An angle in a segment is one formed by two 
straight lines drawn from any point in the arc 
of the segment to the extremities of its chord. 

We have seen in Theorem^ that a circle may be drawn 
through any three points not in a straight line. But it is 
only under certain conditions that a circle can be drawn 
through more than three points. 

Definition. If four or more points are so placed that a 
circle may be drawn through them, they ar^ said to be 
concyclic. 





160 GEOMETRY 

EXERCISES t 
(Miscellaneous) 

1. All circles which pass through a fixed point, and have their 
centres on a given straight line, pass also through a second fixed point. 

2. If two circles which intersect are cut by a straight lino 
parallel to the common chord, shew that the parts of it intercepted""' 
between the circumferences are equal. 

3. If two circles cut one another, any two parallel straight lines 
drawn through the points of intersection to cut the circles are equal. 

4. If two circles cut one another, any two straight lines drawn 
through a point of section, making equal angles with the common 
chord, and terminated by the circumferences, are equal. 

5. Two circles of diameters 74 and 40 inches have a common 
chord 2 feet in length ; find the distance between their centres. 

6. Draw two circles of radii 10" and 1- 7", and with their centres 
2-1" apart. Find bj'^ calculation, and by measurement, the length 
of the common chord, and its distance from the two centres. 

7. Find the greatest and least straight linos which have one ex- 
tremity on each of two given non-intersecting circles. 

8. If from any point on the circumference of a circle straight 
lines are drawn to the circumference, the greatest is that which 
passes through the centre ; and of anj'^ two such lines the greater 
is that which subtends the greater angle at the centre. 

9. Of all straight lines drawn through a point of intersection of 
two circles and terminated by the circumferences, the greatest is that 
which is parallel to the line of centres. 

^"' 10. // froin any internal point, not the centre, straight lines arc 

j^jj* drawn to the circumference of a circle, then the greatest is that which 

\ . passes through the centre, and the least is the remaining part of that 

\^ diameter; and of any other two such lines the greater is that which 

subtends the greater angle at the centre. 

11. If from any external point straight lines are drawn to the 
circumference of a circle, the greatest is that which passes through the 
centre, and the least is that which when produced passes through the 
centre; and of any other two such lines, the greater is that which sub- 
tends the greater angle at the centre. 



ANGLE PROPERTIES 161 

EXERCISES ON THEOREM 39 

1. Prove thai the angle in a semi-circle is a right angle. 

2. The angle in, a segmeid of a circle greater than a semi-circle is 
an acute angle. 

3. The angle in a segment of a circle less than a semi-circle is an 
obtuse angle. 

4. A circle described on the hypotenuse of a right-angled triangle 
as diameter, passes through the opposite angular point. 

5. Two circles intersect at A and B ; and through A two diam- 
eters A P, AQ are drawn, one in each circle ; shew that the points 
P, B, Q are coUinear. 

6. A circle is described on one of the equal sides of an isosceles 
triangle as diameter. Shew that it passes through the middle point 
of the base. 

7. Circles described on any two sides of a triangle as diameters 
intersect on the third side, or the third side produced. 

8. A straight rod of given length slides between two straight 
rulers placed at right angles to one another ; find the locus of its 
middle point. 

9. Fi7id the locus of the middle points of chords of a circle drawn 
through a fixed point. Distinguish between the cases when the given 
point is within, on, or without the circumference. 

10. If two chords intersect within a circle, they form an angle equal 
to that at the centre, subtended by half the sum of the arcs they cut off. 

11. If two chords intersect without a circle, they form an angle equal 
to that at the centre subtended by half the difference of the arcs they cut 
off. 

12. The sum of the arcs cut off by two chords of a circle at right 
angles to one another is equal to the semi-circumference. 

Definition. A sector of a circle is a figure / \ 

])Oiinded by two radii and the arc intercepted 
between them. 




L 



162 



GEOMETRY 



A 9 



/ Theorem 40. [Euclid III. 21] 

Angles in the same segment of a circle arc eqiicCl. 

A 





Pig.i. 



Fig. 2. 



Let BAC, BDC be angles in tiie same segment BADC of a 
circle, whose centre is 0. 

It is required to prove that the Z BAC = the L BDC. 

Join BO, OC. 

Proof. Because the Z BOC is at the centre, and the 
Z BAC at the O"^, standing on the same arc BC, 

.-. the Z BOC - twice the Z BAC. Thcor. 39. 

Similarly the Z BOC = twice the Z BDC. 

.: the Z BAC = the Z BDC. q.e.d. 



Note. The p:ivon s<>fr"icn( may he frroater than a .semi-circlo as 
in Fig. 1, or less than a senn-circle as in Fig. 2 ; in the latter ease the 
angle BOC will be reflex. But by virtue of the e.xtension of Theorem 
."^9 given on page 159, the above proof applies equally to both 
figures. 




ANGLE PROPERTIES 163 

CONVERSE OF THEOREM 40 

Equal angles standing on the same base, and on the same side of it, 
have their vertices on an arc of a circle, of which the given base is the 
chord. 

Let BAC, BDC be two equal angles standing: 
on the same base Bd, and on the same side of it. 

It is required to prove that A and D lie on an arc 
of a circle having BC as its chord. 

Let ABC be the circle which passes through 
the three points A, B, C ; and suppose it cuts 
BD or BD produced at the point E. 

Join EC. 

Proof. Then the Z BAC = the Z BEC in the same segment. 
But, by hypothesis, the A BAC = the ^ BDC; 

:. the Z BEC = the Z BDC; 

which is impossible unless E coincides with D ; 

.'. the circle through B, A, C must pass through D. 

Corollary. The locus of the vertices of triangles drawn on the 
same side of a given base, and ivith equal vertical angles, is an arc of a 
circle. 

EXERCISES ON THEOREM 40 

1. In Fig. 1, if the angle BDC is 74°, find the number of degrees 
in each of the angles BAC, BOC, OBC. 

2. In Fig. 2, let BD and CA intersect at A'. If the angle DXC 
= 40°. and the angle XCD = 25°, #nd the number of degrees in the 

angle BAC and in the reflex angle BOC. 

3. In Fig. 1. if the angles CBD, BCD are respectively 4.3° and 
82°, find the number of degrees in the angles BAC, OBD, OCD. 

4. Shew that in Fig. 2 the angle OBC is always less than the 
angle BAC by a right angle. 

[For further Exercises on Theorem 40 see page 166.] 



164 GEOMETRY 

^ Theorem 41. [Euclid III. 22] 

The opposite angles of any quadrilateral inscribed in a circle 
are together equal to two right angles. 



D 




Let A BCD be a quadrilateral inscribed in the O ABC. 
It is required to prove that 

(i) the A ADC, ABC together = two rt. angles. 
(ii) the A BAD, BCD together = two rt. angles. 
Suppose is the centre of the circle. 

JoinOA, OC. 
Proof. Since the Z ADC at the O'*' = half the Z AOC 
at the centre, standing on the same arc ABC ; 
and the Z ABC at the O" = half the reflex Z AOC at the 
centre, standing on the same arc ADC ; 
.-. the A ^DC, A5C together = half the sum of the Z AOC 
and the reflex Z AOC. 

But the latter angles make uj:) four rt. angles. * 

.-. the 4 ADC, ABC together = two rt. angles. 
Similarly the A BAD, BCD togethcM- = two rt. angles. 

Q.E.D. 

Note. Tho results of Thoorems 40 and 41 .shoiild bo carpfully 
compared. From Theoroni 40 we learn that anples in the .sainc seg- 
ment are equal. From Theorem 41 we learn that anples in conju- 
gate s(>{jment.s are .fupplcmentnri/. 

Definition. A (luadrilateral is called cyclic \\\\cn a ciicle 
can be drawn through its four vertices. 




ANGLE PROPERTIES lG-5 



CONVERSE OF THEOREM 41 

// a pair of opposite angles of a quadrilateral are supplementary , 
its vertices are concyclic. 

Let A BCD be a quadrilateral in which the 
opposite angles at B and D are supplementary-. 

It is required to prove that the points A, B, C, D 
are concyclic. 

Let ABC be the circle which passes through 
the three points A, B, C ; and suppose it cuts 
AD or AD produced in the point E. 
Join EC. 

Proof. Then since ABCE is a cyclic quadrilateral, 

/. the Z AEC is the supplement of the Z ABC. 
But, by hypothesis, the Z ^DCis the supplement of the Z ABC; 
:. the Z. AEC = the Z ADC; 
which is impossible unless E coincides with D. 
.'. the circle which passes through A, B, C must pass through D; 
that is. A, B, C, D are concyclic. q.e.d. 



EXERCISES ON THEOREISI 41 

1. In a circle of 1-6" radius inscribe a quadrilateral A BCD, 
making the angle ABC equal to 126°. Measure the remaining 
angles, and hence verify in this case that opposite angles are sup- 
plementary. 

2. Prove Theorem 41 by the aid of Theorems 40 and 16, after 
first joining the opposite vertices of Uie quadrilateral. 

.3. If a circle can be described about a parallelogram, the 
parallelogram must be rectangular. 

4. ABC is an isosceles triangle, and XY is drawn parallel to the 
base BC cutting the sides in X and Y; shew that the four points 
B, C, X, Y lie on a circle. 

5. // one side of a cyclic quadrilateral is produced, the exterior 
angle is equal to the opposite interior angle of the quadrilateral. 



166 GEOMETRY 



EXERCISES OX ANGLES IN A CIRCLE 

L P is any point on the are of a segment of whicli A B is tlio 
chord; shew that the sum of the angles PAB, PBA is constant. 

2. PQ and RS are two cliords of a circle intersecting at A' ; 
prove that the triangles PXS, RXQ are equiangular to one another. 

3. Two circles intersect at .1 and B ; and through A any 
straight line PAQ is drawn terminated by the circumferences; 

"-^T&^shew that PQ subtends a constant angle at B. 

4. Two circles intersect at A and B ; and through .4 any two 
straight lines PAQ, XAY are drawn terminated by the circumfer- 
ences; shew that the arcs PA', QY subtend equal angles at B. 

5. P is any point on the arc of a segment whose chord is .1 B ; 
and the angles PAB, PBA are bisected by straight lines which 
intersect at 0. Find the locus of the point 0. 

6. If AB is a fixed chord oj a circle and P any point on one of the 
arcs cut off by it, then the bisector of the angle A P B cuts the conjugate 
arc in the same point for all positions of P. 

7. AB, AC are any two chords of a circle; and P, Q are the 
middle points of the minor arcs cut off by them; if PQ is joined, 
cutting AB in A and AC in 1', shew that .lA = AY. 

8. A triangle ABC is inscribed in a circle, and the bisectors of 
the angles meet the circumference at A, 1', Z. Shew that the angles 
of the triangle XYZ are respectively 

90"- — , 90°-^, 90°-^. 

0. Two circles intersect at A and B; and through these points 
lines are drawn from any point P on the circumference of one of the 
circles; shew that when produced they intercept on the other cir- 
cumference an arc which is constant for all positions of /'. 

10. Tlic straight lines which join the e.Ntri-mities of parallel 
chords in a circle (i) towards the same parts, (ii) towards opposite 
parts, arc equal. 



EXERCISES ON ANGLKS IN A CIRCLE 1G7 

11. Through A, a point of intersoetion ot' two equal circles, two 
straight lines PAQ, XAY are drawn; shew that the chord PA' is 
equal to the chord QY 

12. Through the points of intersection of two circles two parallel 
straight lines are drawn terminated by the circumferences ; shew 
that the straight lines which join their extremities towards the same 
jjarts are equal. 

13. Two equal circles intersect at A and B ; and through .1 any 
straight line PAQ is drawn terminated by the circumferences ; shew 
that BP = BQ. 

14. ABC is an isosceles triangle inscribed in a circle, and the 
l)isectors of the base angles meet the circumference at X and Y. 
Shew that the figure BXA YC must have four of its sides equal. 

What relation must subsist among the angles of the triangle 
ABC, in order that the figm-e BXAYC may be equilateral? 

15. ABC D is a cyclic quadrilateral, and the opposite sides AB, 
DC are produced to meet at P, and CB, DA to meet at Q; if the 
circles circumscribed about the triangles PBC, QAB intersect at R, 
shew that the points P, R, Q are collinear. 

IG. P, Q, R are the middle points of the sides of a triangle, and X 
is the foot of the perpendicular let fall from one vertex on the opposite 
side; shew that the four points P, Q, R, X are concyclic. 

[See page 64, Ex. 2; also Prob. 10, p. 83.] 

17. Use the preceding exercise to shew that the middle points oftlie 
sides of a triangle and the feet of the perpendiculars let fall from the 
vertices on the opposite sides, are concyclic. 

18. If a series of triangles are ^Kt-wn standing on a fixed base 
and having a given vertical angle, shew that the bisectors of th(> 
vertical angles all pass through a fixed point. 

19. ABC is a triangle inscribed in a circle, and E the middle 
point of the are subtended by BC on the side remote from A ; if 
through E a diameter E D is drawn, shew that the angle DEA is 
half the difference of the angles at B and C. 



L 



168 



GEOMETRY 



TANGENCY 
Definitions and First Principles 

1. A secant of a circle is a straight line of indefinite 
length which cuts the circumference at two points. 

2. If a secant moves in such a wa}' that the two points 
in which it cuts the circle continually approach one another, 
then in the ultimate position when these two points become 
one, the secant becomes a tangent to the circle, and is said to 
touch it at the point at which the two intersections coincide. 
This point is called the point of contact. 

For instance : 

(i) Let a secant cut the circle at the points P 
and Q, and suppose it to recede from the centre, 
moving so as to be always parallel to its original 
position ; then the two points P and Q will 
clearly approach one another and finally coin- 
cide. In the ultimate position when P and 
Q become one point, the straight line b(^ 
comes a tangent to the circle at that point. 

(ii) Let a secant cut the circle at the points 
P and Q, and suppose it to be turned about 
the point P so that while P remains fixed, Q 
moves on the circumference nearer and nearer 
to P. Then the line PQ in its ultimate 
position, when Q coincides with P, is a tan- 
gent at the point P. 

Since a secant can cut a circle at two points only, it is clear 
that a tangent can have only one point in common with the 
circumference, namely the point of contact, at which two 
points of section coincide. Hence we maj' define a tangent 
as follows : 

3. A tangent to a circle is a straight line which meets 
the circumference at one point only ; and though produced 
indefinitely does not cut the circumference. 






Fig. I. 



Fig. 2. 



Fig.3- 



4. Let two circles intersect (as in Fig. 1) in the points 
P and Q, and let one of the circles turn about the point P, 
which remains fixed, in such a way that Q continually ap- 
proaches P. Then in the ultimate position, when Q coincides 
with P (as in Figs. 2 and 3), the circles are said to touch one 
another at P. 

Since two circles cannot intersect in more than two points, 
two circles which touch one another cannot have more than 
one point in common, namely the point of contact at which 
the two points of section coincide. Hence circles are said to 
touch one another when they meet, but do not cut one 
another. 

Note. When each of the circles wliich meet is outside the other, 
as in Fig. 2, they are said to touch one another externally, or to have 
external contact; when one of the circles is within the other, as in 
Fig. 3, the first is said to touch the other internally, or to have in- 
ternal contact with it. 

Inference from Definitions 2 and 4 

If in Fig. 1, TQP is a common chord of two circles one 
of which is made to turn about P, then when Q is brought 
into coincidence with P, the line TP passes through two coin- 
cident points on each circle, as in Figs. 2 and 3, and therefore 
becomes a tangent to each circle. Hence 

Two circles which touch one another have a common tangent at 
their point of coritact. 



5~t-« Fo-o-v^r^.,^ " ^- 



O, 



170 GEOMETRY 



Theorem 42 



H 



T/ie tangent at any point of a circle is perpend icnlar to the 
radius drawn to the point of contact. 




Let PT be a tangent at the point P to a circle whose centre 
isO. 

It is required to prove that PT is perpendicular to the radius 
OP. 

Proof. Take any point Q in PT, and join OQ. 
Then since P J" is a tangent, every point in it except P is 
outside the circle. 

•'. OQ is greater than the radius OP. 

And this is true for everj^ point Q in PT ; 

.'. OP is the shortest distance from to PT. 

Hence OP is perp. to PT. Theor. 12, Cor. 1. 

y.E.D. 

Corollary 1. Since there can be only one j^erpendiculnr 
to OP at the point P, it follows that one and only one tangent 
can he drawn to a circle at a given point on the circumjeretice. 

Corollary 2. Since there can be only one perpendicular 
to PT at the point P, it follows that the perpend icular to a 
tangent at its point of contact pa,sses throiu/h the centre. 

Corollary 3. Since there can be only one perpendicular 
from to the line PT, it follows that the radius drawn perpen- 
dicular to the tangent passoi through the point of contact. 



TANGENCY 171 

Theorem 43 
Twu tangents can he drawn to a circle from an external paint. 

jp 




Let PQR be a circle whose centre is 0, and let T be an 
external point. 

It is required to prove that there can be two tangents drawn to 
the circle from T. 

Join OT, and let TSO be the circle on OT as diameter. 

This circle will cut the O PQR in two points, since T is 
without, and is within, the O PQR. Let P and Q be these 
points. 

Join TP, TQ ; OP, OQ. 

Proof. Xow each of the A TPO, TQO, being in a semi- 
circle, is a rt. angle ; 

.'. TP, TQ are perp. to the radii OP, OQ respectively. 

.•! TP, TQ are tangents at P and Q. Theor. 42. 

Q.E.D. 

Corollary. The two tangents to a circle from an external 
point are equal, and suhteml equal angles at the centre. 
For in the A TPO, TQO, 
[ the A TPO, TQO are right angles, 
because < the hj-potenuse TO is common, 
[ and OP = OQ, being radii ; 
.-. TP = TQ, 
and the Z TOP = the Z TOQ. Theor. 18. 



172 GEOMETRY 

EXERCISES ON THE TANGENT 

{Numerical and Graphical) 

1. Draw two concentric circles with radii 5-0 cm. and 3-0 cm. 
Draw a series of chords of the former to touch the latter. Calcu- 
late and measure their lengths, and account for their being equal. 

2. In a circle of radius 1-0" draw a number of chords each 1-6" 
in length. Shew that they all touch a concentric circle, and find 
its radius. 

3. Find to the nearest millimetre the length of any chord of a 
circle of radius 5-0 cm., which touches a concentric circle of radius 
2-5 cm., and check your work by measurement. 

4. In the figure of Theorem 43, if OF = 5", TO = 13", find the 
length of TP and TQ. Draw the figure (scale 2 cm. to 5"), and meas- 
ure to the nearest degree the angles subtended at O by the tangents. 

5. The tangents from T to a circle whose radius is 0-7" are each 
2-4" in length. Find the distance of T from the centre of the circle. 
Draw the figure and check your result graphically. 

( Theoretical) 

6. The centre of any circle which touches two intersecting straight 
lines must lie on the bisector of the angle between them. 

7. AB and AC are two tangents to a circle whose centre is (); 
shew that AO bisects the chord of contact BC at right angles. 

8. If PQ is joined in tlu^ figure of Theorem 43 shew that the 
angle PTQ is double the angle OPQ. 

9. Two parallel tangents to a circle intercept on .any third tan- 
gent a segment which subtends a right angle at the centre. 

10. The diameter of a circle bisects all chords which are parallel 
to the tangent at either extremity. 

11. Find the locus of the centres of nil circles which touch (i) n 
given straight line at a given ■point, (ii) each of two parallel straight 
lines, (iii) each of two intersecting straight lines. 

12. In any quadrilateral circumscribed about a circle, the sum of 
one pair of opposite sides is equal to the su7n of the other pair. 

State and prove the converse theorem. 

13. If a quadrilateral is described about a circle, the angles sub- 
tended at the centre ljy any two opposite sides are supplementary. 



THE CONTACT OF CIRCLES 173 

1 



Theorem ^ 



H 



If two circles touch one another, the centres ami the point of 
contact are in one straight line. 





Let two circles whose centres arc and Q touch at the 
point P. 

It is required to prove that 0, P, ami Q are in one straight line. 
Join OP, QP. 

Proof. Since the given circles touch at P, they have a 
common tangent at that point. Page 169. 

Suppose PT to touch both circles at P. 
Then since OP and QP are radii drawn to the point of 
contact, 

.-. OP and QP are both perp. to PT ; 
.-. OP and QP are in one st. line. Theor. 2. 

That is, the points 0, P, and Q^re in one st. line, q.e.d. 

Corollaries, (i) If two circles touch externally thedistance 
between their centres is equal to the sum of their radii. 

(ii) If two circles touch internally, the distance between their 
centres is equal to the difference of their radii. 



174 GEOMETRY 

EXERCISES ON THE CONTACT OF CIRCLES 

{Numerical and Graphical) 

1. From centres 2-6" apart draw two circles Avitli radii 1-7" 
and 0-9" respectively. Why and where do these circles touch? 

If circles of the above radii are drawn from centres 0-8" apart, 
prove that they touch. Plow and why does the contact differ from 
that in the former case? 

2. Draw a triangle ABC in which a = 8 em., 6=7 cm., and 
c = 6 cm. From A, B, and C as centres draw circles of radii 2-5 
cm., 3-5 cm., and 4-5 cm. respectively; and shew that these circles 
touch in pau's. 

3. In the triangle ABC, right-angled at C, o =8 cm. and 6=6 
cm. ; and from centre A with radius 7 cm. a circle is drawn. Find 
the radius of a circle drawn from centre B to touch the first circle. 

4. A and B are the centres of two fixed circles which touch in- 
ternalh\ If P is the centre of any circle Avhich touches the larger 
circle internally and the smaller externally, prove that AP + BP 
is constant. 

If the fixed circles have radii oO cm. and 30 cm. respectively, 
verify the general result by taldng different positions for P. 

5. ^ J5 is a line 4" in length, and C is its middle point. On .4 B, 
AC, CB semi-circles are described. Shew that if a circle is inscribed 
in the space enclosed by the three semi-circles its radius must be |". 

( Theoretical) 

6. A straight li^ie is drawti through the point of contact of two 
circles whose centres are A and B, cutting the circumferences at P and (j 
respectively; shew that the radii AP and BQ are parallel. 

7. Two circles touch externally, and through the point of con- 
tact a straight line is drawn terminated by the circumferences ; 
shew that the tangents at its extremities are parallel. 

8. Find the locus of the centres of all circles which touch a 
given circle (i) at a given point ; (ii) and are of a given radius. 

0. From a given point as centre describe a circle to touch a 
given circle. ITow many solutions will there be? 

10. Describe a circle of radius a to touch a gi\ en circle of radius 
b at a given point. How many solutions will there be? 



THE COXTArT OF CIRCLES 



175 



'i'iiK(jHi;.\i t-l/'^IKuclid III. ;}2] 

The angles made by a tangent to a circle with a chord drawn 
from the point of contact are respectively equal to the angles in 
the alternate segments of the circle. 

A 




Let EF touch the O ABC at B, and let BD be a chord 
drawn from B, the point of contact. 
It is required to prove that 

(i) the Z FBD = the angle in the alternate segment BAD ; 
(ii) the Z EBD = the angle in the alternate segment BCD. 
Let BA be the diameter through B, and C any point in the 
arc of the segment which does not contain A. 
Join AD, DC, CB. 
Proof. Because the Z ADB in a semi-circle is a rt. angle, 
.•. the A DBA, BAD together = a rt. angle. 
But since EBF is a tangent, and BA a diameter, 
.*. the Z FBA is a rt. angle. 
.-. the Z FBA = the A DBA, BAD together. 
Take away the common Z DBA, 
then the Z FBD = the Z BATT, in the alternate segment. 
Again because ABCD is a cyclic quadrilateral, 
/. the Z BCD = the supplement of the Z BAD 
= the supplement of the Z FBD 
= the Z EBD ; 
.*. the Z EBD = the Z BCD, in the alternate segment. 

Q.E.D. 



1 76 GEOMETRY 

EXERCISES ON THEOREM 45 

1. In the figure of Theorem 45, if the A FBD = 72^ write 
down the values of the A BAD, BCD, EBD. 

2. Use this theorem to shew that tangents to a circle from an 
external point are equal. 

3. Through A, the point of contact of two circles, chords APQ, 
AXY are drawn; shew that PA' and QY are parallel. 

Prove this (i) for internal, (ii) for external contact. 

4. ylB is the common chord of two circles, one of which passes 
through 0, the centre of the other; prove that OA bisects the angle 
between the common chord and the tangent to the first circle at A. 

5. Two circles intersect at A and B ; and through P, any point 
on one of them, straight lines P AC, FBD are drawn to cut the other 
at C and D; shew that CD is parallel to the tangent at P. 

0. If from the point of contact of a tangent to a circle a chord 
is drawn, the perpendiculars dropped on the tangent and chord 
from the middle point of either arc cut off by the chord are equal. 

7. Deduce Theorem 44 from the property that the line of cc7ilres 
bisects a common chord at right angles. 

8. Deduce Theorem 45 from Ex. 5, page 165. 

9. Deduce Theorem 42 from Theorem 39. 



PROBLEMS ON CIRCLES 177 



PROBLEMS 

Geometrical Analysis 

Hitherto the Propositions of this text-])ook havo boon 
arranged Synthetically, that is to say, by building up knouii 
results in order to obtain a new result. 

But this arrangement, though convincing as an argument, 
in most cases affords httle clue as to the way in which the 
construction or proof was discovered. We therefore draw the 
student's attention to the following hints. 

In attempting to solve a problem begin bj^ assuming the 
required result ; then by working backwards, trace the conse- 
quences of the assumption, and try to ascertain its depend- 
ence on some condition or known theorem which suggests the 
necessary construction. If this attempt is successful, the 
steps of the argument may in general be re-arranged in 
reverse order, and the construction and proof presented in a 
synthetic form. 

This unravelling of the conditions of a proposition in order 
to trace it back to some earlier principle on which it depends 
is called geometrical analysis : it is the natural way of attack- 
ing the harder types of exercises^ and it is especially useful in 
solving problems. 

Although the above directions do not amount to a method, 
they often furnish a very effective mode of searching for a 
suggestion. The approach by analysis will be illustrated in 
some of the follov\^ing problems. [See Problems 24, 29, 30.] 



178 



GEOMETRY 



Given a circle, or an arc of a circle, to find its centre. 
Let ABC be an arc of a circlo whose 
centre is to be found. 

Construction. Take two chords .45, 
BC , and bisect them at riglit angles by 
the Unes DE, FG, meeting at 0. 

Proh. 2. 

Then is the required centre. 

Proof. Every point in DE is equidis- 

tant from A and B. Proh. 14. 

And every point in FG is equidistant from B and C. 
:. is equidistant from A, B, and ('. 
.'. is the centre of the circle ABC. Theor. 36. 




Problem 22 

To bisect a given arc. 

Let ADB be the given arc to be bisected. 
Construction. Join AB, and bisect it at 
right angles by CD meeting the arc at D. 

Proh. 2. 
Then the arc is bisected at D. 
Proof. Join DA, DB. X 

Then every point on CD is equidistant from A and B ; 

Proh. 14. 
.-. DA = DB ; 
.: the Z DBA = the Z DAB; Theor. (i. 

.•. the arcs, which subtend these angles at the O*"", are ('(iuai ; 
that is, the arc DA = the arc DH. 




TANGENTS 179 



Q.^ 



Problem ^ 

To draw n tangent to a circle from a given external point. 




Let PQR be the given circle, with its centre at ; and let 
T he the point from which a tangent is to be drawn. 

Construction. Join TO, and on it describe a semi-circle 
TPO to cut the circle at P. 

Join TP. 

Then TP is the required tangent. 

Proof. Join OP. 

Then since the Z TPO, being in a semi-circle, is a rt. angle, 
.'. TP is at right angles to the radius OP. 

.: TP is a tangent at P. Theor. 42. 

Since the semi-circle may be described on either side of 
TO, a second tangent TQ can be drawn from T, as shewn in 
the figure. 



180 GEOMETRY 

Problem 2| 
To draw a common tangent to two circles. 
D 

E 




Let A be the centre of the greater circle, and a its radius ; 
and let B be the centre of the smaller circle, and h its radius. 

Analysis. Suppose DE to touch the circles at D and E. 
Then the radii AD, BE being perp. to DE, are parallel. 

Now if BC were drawn pai-' to DE, then the fig. DB would 
be a rectangle, so that CD = BE = b. 

And ii AD, BE are on the same side of AB, 

then AC = a — h, and the Z ACB is a rt. angle. 

These hints enable us to draw BC first, and thus lead to the 
following construction. 

Construction. With centre A, and radius. equal to the 
difference of the radii of the given circles, describe a circle 
and draw BC to touch it. 

Join AC, and produce it to meet the circle (4) at D. 

Through B draw the radius BE par' to AD and in the same 
sense. Join DE. 

Then DE is a common tangent to the given circles. 

Obs. Since two tangents, such as BC, can in general be 
drawn from B to the circle of construction, this method will 
furnish two common tangents to the given circles. These 
are called the direct common tangents. 



COMMON TANGENTS 181 

Pkoblem 24. (Continued) 




Again, if the circles are external to one another two more 
common tangents may be drawn. 

Analysis. In this case we may suppose DE to touch the 
circles at D and E so that the radii AD, BE fall on opposite 
fiides of AB. 

Then BC, drawn pai-' to the supposed common tangent 
DE, would meet AD produced at C ; and we should now 
have 
AC = AD -{- DC = a -\- b ; and the Z ACB is a rt. angle. 

Hence the following construction. 

Construction. With centre A, and radius equal to the mm 
of the radii of the given circles, describe a circle, and draw 
BC to touch it. 

Then proceed as in the first case, but draw BE in the sense 
opposite to AD. ~ 

Ohs. As before, two tangents may be drawn from B to 
the circle of construction ; hence two common tangents may 
be thus drawn to the given circles. These are called the 
transverse common tangents. 

[We leave as an exercise to the student the arrangement of the 
prof)f in synthetic- form.] 



182 GEOMETRY 

EXERCISES ON COMMON TANGENTS 

(Nutnerical and Graphical) 

1. How many common tangents can be drawn (i) when the given 
circles intersect; (ii) when they have external contact; (iii) when 
they have internal contact? 

Illustrate your answer by drawing two circles of radii 1-4" and 
1-0" respectively, (i) with 1-0" between the centres; (ii) witli 
2 -4" between the centres; (iii) w-ithO-4" between the centres; 
(iv) Avith 3-0" between the centres. 

Draw the common tangents in each case, and note where the 
general construction fails, or is modified. 

2. Draw two circles with radii 20" and OS", placing their 
centres 2-0" apart. Draw the common tangents, and find their 
lengths between the points of contact, l)Olh by calculation and l)y 
measurement. 

3. Draw all the common tangents to two circles whose centres 
are 1-8" apart and whose radii are 0-0" and 1-2" respectively. 
Calculate and measure the length of the direct common tangents. 

4. Two circles of radii 1-7" and 1-0" have their centres 2-1" 
apart. Draw their common tangents and find their lengths. Also 
find the length of the common chord. Produce the common chord 
and shew by measurement that it bisects the common tangents. 

.5. Draw two circles with radii 1-6" and 0-8" and with their 
centres 30" apart. Draw all their common tangents. 

6. Draw the direct common tangents to two equal cinles. 

( Theoretical) 

7. If the two direct, or the two transverse, common tangents 
are drawn to two circles, the parts of the tangents intercepted be- 
tween the points of contact are equal. 

8. If four common tangents are drawn to two circles external 
to one another, shew that the two direct, and also the two trans- 
verse, tangents intersect on the line of centres. 

9. Two given circles have external contact at A, and a direct 
common tangent is drawn to touch them at P and Q; shew that PQ 
.subtends a right angle at the point .1. 



THE CONSTRUCTION OF CIRCLES 183 

On the Construction of Circles 

In order to draw a circle we must know (i) the position of 
the centre, (ii) the length of the radius. 

(i) To find the position of the centre, two conditions are 
needed, each giving a locus on which the centre must lie ; so 
that the one or more points in which the two loci intersect 
are possible positions of the required centre, as explained on 
page 93. 

(ii) The position of the centre being thus fixed, the radius 
is determined if we know (or can find) any point on the 
circumference. 

Hence to draw a circle three independent data are required. 

For example, we may draw a circle if we are given (i) three 
points on the circumference ; or (ii) three tangent lines ; or (iii) one 
point on the circumference, one tangent, and its point of contact. 

It will however often happen that more than one circle can be 
drawn satisfying three given conditions. 

Before attempting the constructions of the next Exercise 
the student should make himself familiar with the following 
loci. 

(i) The locus of the centres of circles iphich pass through two . 
given points. 

(ii) The locus of the centres of circles which touch a given 
straight line at a given point. 

(iii) The locus of the centres of circles which tauch a given 
circle at a given point. ^ 

(iv) The locus of the centres of circles which touch a given 
straight line, and have a given radius. 

(v) The locus of the centres of circles which touch a given 
circle, and have a given radius. 

(vi) The locus of the centres of circles v:hich touch two given 
straight lines. 



184 GEOMETRY 

EXERCISES 

1. Draw a circle to pass through three given points. 

2. If a circle touches a given line PQ at a point A, on what line 
must its centre lie? 

If a circle passes through two given points A and B, on what line 
must its centre he? 

Hence draw a circle to touch a straight line PQ at the point A, 
and to pass through another given point B. 

3. If a circle touches a given circle whose centre is C at the point 
-1, on what line must its centre lie? 

Draw a circle to touch the given circle (C) at the point A, and 
to pass through a given point B. 

4. A point P is 4-5 em. distant from a straight line AB. Draw 
two circles of radius 3- 2 cm. to pass through P and to touch A B. 

5. Given two circles of radius 3-0 em. and 2-0 cm. respectively, 
their centres being 60 cm. apart; draw a circle of radius 3-5 cm. 
to touch each of the given circles externally. 

How many solutions will there be? What is the radius of the 
smallest circle that touches each of the given circles externally? 

6. If a circle touches two straight lines AO, OB, on what lino 
must its centre lie? 

Draw OA, OB, making an angle of 76°, and describe a circle of 
radius 1-2" to touch both lines. 

7. Given a circle of radius 3- 5 cm., ^vith its centre ")0 cm. from 
a given straight line AB; draw two circles of radius 2-5 cm. to 
touch the given circle and the line AB. 

8. Devise a construction for drawing a circle to touch each of 
two parallel straight lines and a transversal. 

Shew that two such circles can be drawn, and lluit tluy are equal. 

9. Describe a circle to touch a given circle, and also to touch a 
given straight line at a given point. 

10. Describe a circle to touch a given straight line, and to touch 
a given circle at a given point. 

11. Shew how to draw a circle to touch each of three given 
straight lines of which no two are paralh^l. 

How manv such circles can be drawn? 



PROBLEMS 185 

, ^H • 

Problem,^ 

On a given straight line to describe a segment of a circle which 
shall contain an angle equal to a given angle. 





Let AB be the given st. line, and C the given angle. 
It is required to describe on AB a segment of a circle contain- 
ing an angle equal to C. 

Construction. At A in BA, make the Z BAD equal to the 
Z C. 

From A chaw .46"perp. to AD. 

Bisect AB at rt. angles by FG, meeting AG' in G. Prob. 2. 

Proof. Join GB. 

Now every point in FG is equidistant from A and B ; 

Prob. 14. 
.-. GA = GB. 

With centre G, and radius GA, draw a circle, which must 
pass through B, and touch AD^ A. 

Then the segment AHB, alternate to the Z BAD, contains 
an angle equal to C. Theor. 45. 

Note. In the particular case when the given angle is a rt. angle, 
the segment required will be the semi-circle on AB as diameter. 
[Theorem 39.] 



18G GEOMETRY 

Corollary. To cut off from a given circle a segment con- 
taining a given angle, it is enough to draw a tangent to the circle, 
and from the point of contact to draw a chord making with the 
tangent an angle equal to the given angle. 

It was proved on page 163 that 
The locus of the vertices of triangles which stand on the same base 
and have a given vertical angle, is the arc of the segment standing 
on this base, and containing an angle equal to the given angle. 

The following Problems are derived from this result by the 
Method of Intersection of Loci [page 93]. 

EXERCISES 

1. Describe a triangle on a given base having a given vertical angle 
and having its vertex on a given straight line. 

2. Construct a triangle having given the base, the vertical angle, and 

(i) one other side. 
(ii) the altitude. 

(iii) the length of the median which bisects the base. 
(iv) the foot of the perpendicular from the vertex to the base. 

3. Construct a triangle having given the base, the vertical angle, and 
the point at which the base is cut by the bisector of the vertical angle. 

[Let AB be the base, A' the given point in it, and A' the given 
angle. On AB describe a .segment of a cirele containing an angle 
equal to A'; complete the C® Jiy drawing the arc APB. Bisect 
the arc APB at P : join PA', and produce it to meet the O*^* at ('. 
Then ABC is the required triangle.] 

4. Construct a triangle having given the base, the vertical angle, and 
the sum of the remaining sides. 

(Let AB be the given base, A' the given angle, and // a line equal 
to the sum of the sides. On AB describe a segment containing an 
angle equal to A', also another segment containing an angle equal to 
half the Z K. With centre A, and radius //, describe a circle cut- 
ting the arc of the latter segment at X and }'. Join .1 A' (or .1 }') 
cutting the arc of the first segmcMit at C Then .1 BC is (he required 
triangle.) 

5. Construct a triangle having given the base, the vertical nugic, and 
the difference of the remaining sides. 



CIRCLES AND POLY(J()XS 187 



CIRCLES IX RELATION TO RECTILINEAL 
FIGURES 

Definitions 

L A Polygon is a rectilineal fig;ure bounded by more than 
four sides. 

A Polygon of five sides is called a Pentagon, 

A Poh'gon of six sides is called a Hexagon, 

A Polygon of seveii sides is called a Heptagon, 

A Polygon of eight sides is called an Octagon, 

A Poh'gon of ten sides is called a Decagon, 

A Polygon of twelve sides is called a Dodecagon, 

A Polygon of fifteen sides is called a Quindecagon. 

2. A Polygon is Regular when all its sides are equal, and 
all its angles are equal. 

3. A rectilineal figure is said to be in- 
scribed in a circle, when all its angular points 
are on the circumference of \hv circle ; and a 
circle is said to be circumscribed about a recti- 
lineal figure, when the circumference of the 
circle passes through all the angular points of 
the figure. 

4. A circle is said to be inscribed in a 
rectilineal figure, when the circumference of 
the circle is touched by each side of the figure ; 
and a rectilineal figure is said to be circum- 
scribed about a circle, when each side of the 
figure is a tangent to the circle. 





188 



GEOMETRY 



Problem 2G 
To circumscribe a circle about a given triangle. 




Let ABC be the triangle, about which a circle is to bo 
drawn. 

Construction. Bisect AB and AC at rt. angles by DS and 
ES, meeting at S. Prob. 2. 

Then S is^thc centre of the required circle. 

Proof. Now every point in DS is equidistant from A 
and B ; ' Prob. 14. 

and every point in ES^is equidistant from A and C ; 
.■. *S is equidistant from A, B, and C. 
With centre S, and radius SA describe a circle; this will 
pass through B and C, and is, therefore, the required circum- 
circle. 

06.S. It will be found that if the given triangle is acute- 
angled, the centre of the circum-circle falls within it : if it 
is a right-angled triangle, the centre falls on the hypotenuse : 
if it is an obtuse-angled triangle, the centre falls without the 
triangle. 

Note. From page 01 it is seen that if S is joined to tlie middle 
point of BC, then the joining line is perpendicular to BC. 

Hence the perpendiculars drawn to the sides of a triangle from their 
middle points are concurrent,' the point of intersection being the centre 
of the circle circumsrrihcd about the Irinnijh. 



PROBLEMS ON TRIANGLES AND CIRCLES 189 

Proble]\I'::3?^ ^ c? 

To inscribe a circle in a given triangle. 
A 

•/- 




L 



C 

Let ABC be the triangle, in which a circle is to be in- 
scribed. 

Construction. Bisect the A ABC, ACB by the st. lines 
BI, CI, which intersect at /. Prob. 1. 

Then I is the centre of the required circle. 
Proof. From I draw ID, IE, IF perp. to BC, CA, AB. 
Then every point in BI is equidistant from BC, BA : Prob. 15. 
.-. ID = IF. 
And every point in CI is equidistant from CB, CA ; 
.-. ID =. IE. 
.: ID, IE, IF are all equal. 
With centre / and radius ID draw a circle ; 
this will pass through the points E and F. 
Also the cii'cle will touch the sides BC, CA, AB, 
because the angles at D, E, F are right angles. 
.-. the O DEF is inscribed in the A ABC. 

Note. From II., p. 97 and Problem 27 it follows that 
The bisectors of the angles of a triangle are concurrent, the point of 
intersection being the centre of the inscribed circle. 

Definition. A circle which touches one side of a triangle 
and the other two sides produced is called an escribed circle 
of the triangle. 



1 90 GEOMETRY 

Problem 2S ^T 

To draw an escfiihed circle of a cjiven triangle. 




Let ABC be the given triangle of which the sides AB, AC 
are produced to D and E. 

It is required to describe a circle touching BC, and BD, CE. 
Construction. Bisect the A CBD, BCE b}- the st. hnes 
BIu CIi which intersect at /i. 

Then 7i is the centre of the required cii-cle. 
Proof. From /i draw I.F, 1,0, 1,H pcrp. to AD, BC, AE. 
Tlion ever}' point in BIi is equicHstant from BD, BC] Proh. 15. 

.-. /,F = 7,(7. 
Simihirly LG = I^H. 
.'. LF, liG, IJI are all equal. ■ 
With centre 7i and radius JiF descri])e a circle ; 
this will pass through the points (/ and //. 
Also the circle will touch AD, BC, and AE, 
l)ecause the angles at F, G, II are rt. angles. 
.*. the O FGII is an escribed circle of the A ABC. 

Note 1. It i.s clear that every triangle has three escribed <'irclt>s. 

Their centres are known as the Ex-centres. 

Note 2. From II a, page 97 and Problem 28 it follows that 
The bisectors of two exterior angles of a triangle and of the third 

angle are concurrent, the point of intersection being an ex-centre. 



PROBLEMS ON CIRCLES AND TRLINGLES 191 

Problem 29 

In a given circle to inscribe a triangle equiangular iu a given 
triangle. 

J3 




E F 



Let ABC be the given circle, and DEF the given triangle. 
Analysis. A triangle ABC, equiangular to the A DEF, is 
inscribed in the circle, if from any point .4 on the O*^^ two 
choirs AB.AC pan be so placed that, on joining BC, the 
Z E, anct th€^ Zt^=' tlte Z F; for then the Z A = the 
Z B = the Z D. Theor. IG. 

Now the Z B, in the segment ABC, suggests the equal 
angle between the chord AC and the tangent at its extremitj'- 
{Theor. 49) ; so that, if at A we draw the tangent GAH, 
then the Z HAC = the Z E ; 
and similarly, the Z GAB = the Z F. 
Reversing these steps, we have the following construction. 
Construction. At any point A on the O"^ of the O ABC 
draw the tangent GAH. Proh. 23. 

At A make the Z GAB equal to the Z F, 
and make the Z HAC equal to the Z E. 

Join BC. 
Then A5C is the required triangle. 

Note. In drawing the fig-ure on a larger scale the student should 
shew the construetion lines for the tangent GAH and for the angles 
GAB, HAC. A similar remark applico to the next Problem. 



192 



GEOMETRY 



Problem 30 

About a given circle to circumscribe a triangle equiangular to 
a given triangle. 




D 



M B 



N G E F H 



Let ABC be the given circle, and DEF the given triangle. 

Analysis. Suppose LMN to be a circumscribed triangle in 
which the Z iV/ = the ^ E, the Z A' = the Z F, and conse- 
quently, the Z L = Z D. 

Let us consider the radii KA, KB, KC, drawn to the 
points of contact of the sides; for the tangents LM, MN, 
NL could be drawn if we knew the relative positions of KA, 
KB, KC, that is, if we knew the A BKA, BKC. 

Now from the quad' BKAM, since the A B and A are 

rt A 

the Z BKA = 180° - M = 180° - E ; 
similarly the A B KC = 180° - A^ = 180° - F. 
Hence we have the following construction. 

Construction. Produce EF both ways to G and H. 
Find K the centre of the O ABC, 
and draw any radius KB. 
At A' make the Z BKA equal to the Z DEO ; 
and make the Z BKC equal to the Z DFH. 
Through A , B, C draw LM, MN,NL pcM p. to AM , KH, KC. 
Then LMN is th(> retpiired triangle. 

[The student should uow arrange the i)roof syntljetically.I 



PROBLEMS ON CIRCLES AND TRIANGLES 193 

EXERCISES 

On Circles and Triangles 

(Inscriptions and Circumscriptions) 

1. In a circle of radius 5 cm. inscribe an equilateral triangle; 
and about the same circle circumscribe a second equilateral triangle. 
In each case- state and justify your construction. 

2. Draw an equilateral triangle on a side of 8 cm., and find l\v 
calculation and measurement (to the nearest millimetre) the radii 
of the inscribed, circumscribed, and escribed circles. 

Why are the latter radii double and treble of the first? 

3. Draw triangles from the following data : 

(i) a = 2-5", B = 66°, C =50°; 
(ii) a = 2-o", B=72°, C = 44° ; 
(iii) a = 2-5", B = 41°, C = 23°. 
Circumscribe a circle about each triangle, and measure the radii 
to the nearest hundredth of an inch. Account for the results being 
the same, by comparing the vertical angles. 

4. In a circle of radius 4 cm. inscribe an equilateral triangle. 
Calculate and measure its side to the nearest millimetre. 

Find the area of the inscribed equilateral triangle, and shew that 
it is one quarter of the circumscribed equilateral triangle. 

5. In the triangle ABC, if I is the centre, and r the length of the 
radius of the in-circle, shew that 

AlBC^^ar; AlCA=^br; AlAB = ^cr. 
Hence prove that A ABC ^ ^{a + b + c)r. 

6. If ?i is the radius of the ex-circle opposite to A, prove that 

A ABC = Ub + c - a)ri. 
If a = 5 em., 6=4 cm., c = 3 em., verify by measurement the re- 
sults of Ex. 5 and of this exercise. ^ 

7. Find by measurement the circum-radius of the triangle ABC 
in which a = 6-3 cm., b = 3-0 cm., and c = 5-1 cm. 

Draw and measure the perpendiculars from A, B, C to the oppo- 
site sides. If their lengths are represented by pu p2, pz, verify the 
following statement : 

circum-radius = ^L = S± ^ 11^ . 
2pi 2p2 2p3 



194 GEOMETRY 

EXERCISES 

On Circles and Squares 

{I nscriplious and Circumscriptions) 

1. Draw a circle of radius lo", and find a construction for in- 
scribing a square in it. 

Calculate the length of the side to the nearest hundredth of an 
inch, and verify by measurement. 

Find the area of the inscribed square. 

2. Circumscribe a square about a circle of radius 1-5", shewing 
all lines of construction. 

Prove that the area of the square circumscribed about a circle is 
double that of the inscribed square^ 

3. Draw a square on a side of 7-5 cm., and state a construction 
for inscribing a circle in it. 

Justify your construction by considerations of symmetry. 

4. Circumscribe a cix'cle about a square whose side is 6 cm. 
Measure the diameter to the nearest millimetre, and test your 

drawing by calculation. 

5. In a circle of radius 1-8" inscribe a rectangle of which one 
side measures 3-0". Find the approximate length of the oth(>r side. 

Of all rectangles inscribed in the circle shew that the square has 
the greatest area. 

6. A square and an equilateral triangle are inscribed in a circle. 
If n and b denote the lengths of their sides, shew that 3a'' = 26-. 

7. .1 BCD is a square inscribed in a circle, and P is any point on 
the arc AD: shew that the side AD subtends at P an angle three 
times as great as that subtended at P by any one of the other sides. 

{Prohlemi<. Stale your construction, mid give a theoretical proof.) 
S. Circumscribe a rhombus about a given circle. 
0. Inscribe a square in a given square A BCD, so that one of its 
angular points shall be at a given jmint -Y in AB. 

10. In a given square inscribe the square of mininniin area. 

11. Describe (i) a circle, (ii) a square about a given rectangle. 

12. Inscribe (i) a circle, (ii) a square in a gi\en quadrant. 



PROBLEMS OX CIRCLES AND POLYGONS 195 
OX CIRCLES AXD REGULAR POLYGONS 

To draw a regular polygon (i) in (ii) about a given circle. 

Let AB, BC, CD, •■• be consecutive ^-— ^^^^ n 

sides of a regular polygon inscribed in y^^ /V5vw 

a circle whose centre is 0. / / ^ 

Then AOB, BOC, COD, ••• are con- [ o/-.---- 

gruent isosceles triangles. And if the \ / \ 

° '-' \ . 360 4 

polygon has n sides, each of the A /■-"--", / 

A AOB, BOC, COD, - = — • a''^^^ 

n 

(i) Thus to inscribe a polygon of n sides in a given circle, 
draw at the centre an angle AOB of this size. This gives 
the length of a side AB; and chords equal to ^15 may now 
be set off round the circumference. The resulting figure 
will clearly be equilateral and equiangular. 

(ii) To circumscribe a polygon of n sides about the circle, 
the points A, B, C, D, ■■■ must be determined as before, and 
tangents drawn to the circle at these points. The resulting 
figure may readily be proved equilateral and equiangular. 

Note. This method gives a strict geometrical construction only 
when the angle AOB can be drawn Avith ruler and compasses. 

EXERCISES 

1. By strict constructions inscribe in a circle (radius 4 cm.) a 
regular (i) hexagoa ; (ii) octagon ; (iii) dodecagon. 

2. About a circle of radius 1-5" circumscribe a regular (i) 
hexagon ; (ii) octagon. Test the constructions by measurement, 
and justify them by proof. 

3. Compare the sides and also the areas of an equilateral tri- 
angle and a regular hexagon inscribed in any circle. 

4. Using a protractor inscribe a regular heptagon in a circle of 
radius 2". Calculate and measure one angle ; measure a side. 



196 GEOMETRY 

Problem J^ 

To draw a circle (i) in (ii) about a regular pohjgon. 

Let AB, BC, CD, DE,-- be eon- ^ -^ 

seeutive sides of a regular polygon of //^ ^"a\ 

n sides. // \\ 

Bisect the A ABC, BCD by BO, ^11 ^^>: Wo 

CO meeting at 0. \X^/^^ Ix^// 

Then is the centre both of the ^^>C \/// 

inscribed and circumscribed circle. o'^^^^t^^Q 

Outline of Proof. Join OD ; and from the congruent 
A OCB, OCD, shew that OD bisects the Z CDE and that : 
All the bisectors of the angles of the polygon meet at 0. 
(i) Prove that OB = OC = OD = •■• ; from Theorem 0. 

Hence is the circum-centre. 
(ii) Draw OP, OQ, OR, •■■ perp. to AB, BC, CD, ■■■ . 
Prove that OP = OQ = OR = ■■■ ; from the congruent 
AOBP,OBQ, ••• . 

Hence is the in-centre. 

EXERCISES 

1. Draw a regular hexagon on a side of 20''. Draw llie in- 
scribed and eircumserilied circles. Calculate and measure l!u>ii' 

diameters to the nearest hundredth of an inch. 
* 

2. Shew that the area of a regular hexagon inscribed in a circle 

is three-fourths of that of the circumscribed hexagon. 

Find these to the nearest tenth of a sq. cm. (radius 10 cm.). 

3. If AliC is an isosceles triangle inscribed in a circle, having 
each of the angles B and C double of the angle .1 ; sIuav that liC is 
a side of a regular pentagon insc-ribed in the (-ircle. 

4. On a side of 4 cm. construct (without i^rotractor) a ri'gul;;r 
(i) hexagon ; (ii) octagon ; and in e?ioh case find the approximate 
firca of the figure. 



CIRCUMFERENCE AND AREA OF A CIRCLE 197 

THE CIRCUMFERENCE OF A CIRCLE 

By experiment and measurement it is found that the length 
of the circumference of a circle is roughly 3y times the length 
of its diameter: that is to say 

circumference ^ , , 

— - — = 3f nearly; 

diameter 

and it can be proved that this is the same for all circles. 

A more correct value of this ratio is found by theory to be 
3 1416 ; while correct to 7 places of decimals it is 3 1415926. 
Thus the value Z\ (or 3 14^5) is correct to 2 places only. 

The ratio which the circumference of an}" circle bears to its 
diameter is denoted by the Greek letter tt; so that 
circumference = diameter X tt = 2 ;• X tt = 2 Trr ; 
where r denotes the radius of the circle and where ' to tt we 
are to give one of the values 3-|-, 3-1416, or 3-1415926, ac- 
cording to the degree of accuracy required in the final result . 

Note. The theoretical methods by which tt is evaluated to any 
required degree of accuracy cannot be explained at this stage, but 
its value may be easily verified by experiment to two decimal places. 

For example : round a cylinder vn^ap a strip of paper so that the 
ends overlap. At any point in the overlapping area prick a pin 
through both folds. Unwarp and straighten the strip, then meas- 
ure the distance between the pin holes : this gives the circumference. 
Measure the diameter, and di\ade the first result by the second. 

Ex. 1. From these 
data find and record the 
value of TT. 

Find the mean of the 
three results. 

Ex. 2. A fine thread is wound evenly round a cylinder, and it is 
found that the length required for 20 complete turns is 75-4". The 
diameter of the cylinder is 1-2" : find roughly the value of x. 

Ex. 3. A bicycle wheel, 28" in diameter, makes 400 revolutions 
iu tra\'elliug over 977 vards. Hence estimate the value of tt. 



Circumference. 


Diameter. 


Value of tt. 


16 .^em. 

8-8" 
13-5' 


5 • 1 em. 

2-8" 
4-3" 





l'J8 



GEOMETRY 



THE AREA OF A CIRCLE 

_^Let AB he a, side of a polygon of 
n sides circumscribed about a circle 
whose centre is and radius r. Then 
Area of polygon = n-A AOB 

= n-\AB XOD = \-nAB X r 
= \ {'perimeter of polygon) X r ; 
and this is true however many sides 
the polj'^gon may have. A D B 

Now if the number of sides is increased without limit, the 
perimeter and area of the polygon may be made to differ from 
the circumference and area of the circle by quantities smaller 
than any that can be named; hence ultimately 

Area of circle = h ■circumference X r— h-2Trr X r= tt?-. 

ALTERXATIVE METHOD 






Suppose the circle divided into iinj- oven luiiiilx-r of sectors hav- 
ing equal central angles : denote the number of sectors l)^' /;. 

Let the sectors lie placed side by side as in the diagram ; then 
the area of the cin^le = the area of llu^ fig. A BCD. 

Now if the number of sectors be increased, (>ach arc is decreased ; 
so that (i) the outlines AB, CD tend to become straight, and 
(ii) the angles at D and B tend to become rt. angles. 

Thus when n is increased without limit, the fig. A BCD ulti- 
mately becomes a rcrtanylc, whose length is the seini-circiimfcrcncc of 
the circle, and whose breadth is its radius. 

.'. Area of circle = ]• circumference X radius— i-2n-/- X '" = jrr^ 



CIRCUMFERENCE AND AREA OF A CIRCLE 199 



THE AREA OF A SECTOR 



If two radii of a circle make an angle of 1°, they cut off 
(i) an arc whose length = 3^ of the circumference ; 
and (ii) a sector whose area = -^^ of the circle ; 
.'. if the angle AOB contains D degrees, then 

(i) the arc AB = - — 0/ the circumference ; 
360 

(ii) the sector AOB = — of the area of the circle 
^ 360 

= of (h circumference X radius) 

360 •' ^- •' 

= \arcAB X radius. 

THE AREA OF A SEGMENT 

The area of a minor segment Js found by subtracting from 
the corresponding sector the area of the triangle formed by 
the chord and the radii. Thus 

Area of segment ABC = sector AC B — triangle AOB. 

The area of a major segment is most simply found by 
subtracting the area of the corresponding minor segment from 
the area of the circle. 



200 GEOMETRY 



EXERCISES 

[In each case choose the value of it so as to giue a result of the assigned 
degree of nccuracij.] 

1. I-'ind to the nearest millimetre the circumferences of the 
circles whose radii are (i) 4-5 cm. (ii) 100 cm. 

2. Find to the neai*est hundredth of a square inch the areas of 
llio circles whose radii are (i) 2-3". (ii) 10-()". 

3. Find to two places of decimals the circumference and area 
of a circle inscribed in a square whose side is 3-G cm. 

4. In a circle of radius 7-0 cm. a square is described : find to the 
nearest square centimetre the difference between the areas of the 
circle and the square. 

5. Find to the nearest hundredth of a square inch the area of 
the circular ring formed by two concentric circles whose radii are 
5-7" and 4-3". 

6. Shew that the area of a ring I.ying between the circumferences 
of two concentric circles is equal to the area of a circle whose radius 
is the length of a tangent to the inner circle from any point on the 
outer, 

7. A rectangle whose sides are 8-0 cm. and 6-0 cm. is inscribed 
in a circle. Calculate to the nearest tenth of a square centimetre 
the total area of the four segirtents outside the rectangle. 

8. Find to the nearest tenth of an inch the side of a square 
wlioso area is equal to that of a circle of radius 5". 

9. A circular ring is formed by the circumference of two con- 
centric circles. The area of the ring is 22 square inches, and its 
width is 1-0"; taking tr as V- ^"'1 approximately the radii of the 
two circles. 

10. Find to the nearest Imndredth of a square inch the difference 
b(itw(^en the areas of the circumscribed and inscrib(>d circles of an 
equilateral triangle each of whose sides is 4". 



CIRCLES ASSOCIATED WITH A TRIANGLE 201 



EXERCISES 

( Theoretical) 

1. Describe a circle to touch two parallel straight lines and a 
third straight line which meets them. Shew that two such circles 
can be drawn, and that they are equal. 

2. Triangles which have equal bases and equal vertical angles have 
equal circumscribed circles. 

3. If, in a triangle, ABC, I, S, the centres of the inscribed and 
circumscribed circles, and A are collinear, then AB = AC. 

4. The sum of the diameters of the inscribed and circumscribed 
circles of a right-angled triangle is equal to the sum of the sides 
containing the right angle. 

5. If the circle inscribed in the triangle ABC touches the sides 
at D, E, F; shew that the angles of the triangle DEF are respec- 
tively the complements of the halves of the angles A, B, C. 

6. If / is the centre of the inscribed circle and h the centre of 
the escribed circle of the triangle ABC, then /, B, h, C are concycUc. 

7. In any triangle the difference of two sides is equal to the 
difference of the segments into which the third side is di\'ided at the 
point of contact of the inscribed circle. 

8. In the triangle ABC, I and S are the centres of the inscribed 
and circumscribed circles : then 75 subtends at A an angle equal 
to half the difference of the angles at the base of the triangle. 
Also a AD is perpendicular to BC, A I bisects the Z DAS. 

9. The diagonals of a quadrilateral ABCD intersect at : shew 
that the centres of the circles circumscribed about the four triangles 
AOB, BOC, COD, DO A are the vertices of a parallelogi'am. 

10. In any triangle ABC, if /4s the centre of the inscribed 
circle, and if A 7 is produced to meet the circumscribed circle at O, 
O is the centre of the circum-circle of the triangle BIC. 

11. Given the base, altitude, and the radius of the circumscribed 
circle ; construct the triangle. 

12. Thi-ee circles whose centres are A, B, C touch one another 
externally two by two at D, E, F : shew that the inscribed circle of 
the triangle ABC is the circumscribed circle of the triangle DEF. 



202 GEOMETRY 

EXERCISES 

(Loci) 

1. Given the base BC and the vertical angU^ A of a triangle; 
find the locus of the ex-centre opposite A. 

2. Find the locus of the intersection of the bisectors of the angles 
PAB, QBA if A, B are fixed and PA, BQ are constantly parallel. 

3. Find the locus of the middle points of chords of a circle 
which pass through a fixed point (i) within, (ii) on, (iii) without the 
circumference. 

4. Find the locus of the points of contact of tangents drawn 
from a fixed point to a system of concentric circles. 

5. Find the locus of the intersection of straight lines which pass 
through two fixed points on a circle and intercept on its circum- 
ference an arc of constant length. 

6. A and B are two fixed points on the circumference of a circle, 
and PQ is any diameter; if PA,QB cut in X, find the locus of X. 

7. BAC is any triangle described on the fixed base BC and Iuia- 
ing a constant vertical angle ; and BA is produced to P, so thai .1 /' 
is equal to AC; find the locus of P. 

8. AB is a fixed chord of a circle, and AC is a movable chord 
passing through A ; if the parallelogram CB is completed, find llio 
locus of the intersection of its diagonals. 

9. A straight rod PQ slides between two rul(>rs placed at right 
angles to one another, and from its extremities PX, QX an^ drawn 
perpendicular to the rulers; find the locus of A'. 

10. Two circles intersect at A and B, and /' is any point on tlie 
circumference of one of them. If the lines PA, PB cut the oilier 
circle at X and 1', find the locus of the intersection of .1 Y and BX. 

11. Two circles intersect at A and B ; HAK is a fixed straight 
line drawn through A and terminated by the circumferences, and 
P AQ is any other straight line similarly drawn; find the locus of 
the intersection of IW and QK. 



PART IV 

ON PROPORTION 
Definitions and First Principles 

1. The ratio of one magnitude to another of the same kind 
is the relation which the first bears to the second in regard to 
quantity ; this is measured by the fraction which the first is 
of the second. 

Thus if two such magnitudes contain a and b units respectively 

the ratio of the first to the second is expressed by the fraction -• 

b 

The ratio of a to 6 is generally denoted thus, a ib ; and a 

is called the antecedent and b the consequent of the ratio. 

The two magnitudes compared in a ratio must be of the same 
kind; for example, both must be lines, or both angles, or both 
areas. It is clearly impossible to compare the length of a straight 
line wth a magnitude of a different kind, such as the area of a 
triangle. Moreover, a ratio is an abstract fraction. Thus the ratio 
of a line 6 cm. long to a line 8 cm. long is | or f (not f cm.). 

Note. It is not always possible to express two quantities of the 
same kind in terms of a common unit. For instance, if the side of 
a square is 1 inch, the diagonal is V2 inches. But since the nu- 
merical value of V2 cannot be exactly determined (though it can 
be found to any number of decimal figures), the side and diagonal 
cannot be expressed in terms of the same unit. Two such quan- 
tities are said to be incommensurable. But by choosing a suffi- 
ciently small quantity as unit, two incommensurables, such as V2 
inches and 1 inch, may be expressed to any reqmred degree of 
accuracy. Thus, remembering that V2 = 1- 41421- •• , it follows 
that V2 inches and 1 inch may be represented by 
1414 and 1000, roughly, taking x^" as unit; 

14142 and 10000, more nearly, taldng roiffi?" as unit ; and so on. 

203 



204 GEOMETRY 

2. If a point X is taken in a 

straight line AB, or in AB produced, A^ '. 

then X is said to divide AB into the Fig.i. 

two segments AX, XB; the segments 
being in either case the distances of 



the dim ding point X from the extremi- *^*&2. 

ties of the given line AB. 

3. X is said to divide AB internally in Fig. 1, and ex- 
tcrnall}" in Fig. 2. In the first case AB is the sum, and in 
the second the difference, of the segments AX, XB. In 
cither case the ratio in which X divides AB is the ratio 
of the segments AX, XB. 

4. Four magnitudes a, h, x, y are proportionals or in pro- 
portion, when the ratio of the first to the second is equal to 
the ratio of the third to the fourth. 

This is expressed by saying '' a is to h as x is to y ^' ; and 
the proportion is written 

a_x 
b y 

or a -.h = X : y. 

Here a and y are called the extremes, and }> cud x the 
means ; and y is said to be a fourth proportional to n, b, and x. 

In a proportion, terms which are both antecedents or both 
consequents of the ratios arc said to be corresponding terms. 

Note. In a proportion sucli as a:b = x : »/, the niasniludes 
compared in each ratio must bo of the same land, though the mag- 
nitudes of the second ratio need not be of the same kind as those of 
the first. For instance, a and b may denote areas, and x and y lines; 
in which case the proportion asserts that the ratio of the areas is 
the same as the ratio of tlio lines. 



DEFINITIONS AND I'^IRST PRINCIPLES 205 

5. Three magnitudes of the same kmd are said to be pro- 
portionals, when the ratio of the first to the second is equal to 
that of the second to the third. 

Thus a, b, c are proportionals if 

a: b = b: c. 

Here h is called a mean proportional between a and c ; 
and c is called a third proportional to a and b. 

Introductory Theorems 

I. If four magnitudes are proportionals, they are also pro- 
portionals when taken inversely. 

That is, if a :b = x : y, 

then b : a ^ ij : X. 

For, by hypothesis, 7 = - ; hence - = - ; 
by ax 

or b : a — y : X. 

II. If four magnitudes of the same kind are proportionals, 
they are also proportionals when taken alternately. 

That is, if a :b = x : y, 

then a : X = b :y. 

For, bv hypothesis, - = - ; 

.6 y 

multiplying both sides b}' -' 

X 

, a b x^ h 

we have _._ = _._ j 

b X y X 

that is, ^ = ^, 

X y 

or a : X = b : y. 

Note. In this theorem the hypothesis and conclusion taken 
together require that a, b, x and y shall be of the same kind. 



200 



GEOMETRY 



III. If Jour numbers arc proportional, the product of the 
extremes is equal to the product of the means. 

That is, if a : b = c : d, 

then ad = be. 



For, by hypothesis, 



c 
d' 



multiplying each side of this equation by bd, we have 

ad = be. 

Corollary. If a, b, e, d denote the lengths of four straight 
lines in proportion, the rectangle contained by the extremes is 
equal to the rectangle contained by the ineans. 

This is illustrated by the following diagram : 



radj 



(be) 



Similarly if three lines a, b, c arc proportionals, 
that is, if a :b = b : c ; 

then ac = b-. 

Or, the rectangle contained by the extremes is equal in area to 
the square on the mean. 

IV. If there are four mag7iitudcs in proportion, the sum 
{or difference) of the first and second is to the second as the 
sum {or difference) of the third and fourth is to the fourth. 

That is, if a -.b = x -.y ; 

then (i) a -\- b -.b = x -{- y :y ; 

(ii) a — b :b = x — y : y. 



INTRODUCTORY THEOREMS 207 

For by hvpolhcsis, v — '- ', 

b U 

.-. r + 1 = - + 1, or —j — = —^^ ; 
by by 

that is, a -r b:b = X -{- y: y (i) 

This inference is sometimes referred to as componendo. 

Similarly by subtracting 1 from the equal ratios -, -, we 
obtain ^ 

a — b ^ .T — y 
b y 

that is, a — b:b = X — y :y (ii) 

This inference is sometimes referred to as dividendo. 
Corollary. If a:b = x: y, 

then a -{- b : a — b =^ X -\- y : X — y. 

This is obtained by dividing the result of (i) by that of (ii). 

V. In a series of equal ratios {the magnitudes being all of the 
same kind), as any antecedent is to its consequent so is the sum 
of the antecedents to the sum of the consequents. 

Let each of the equal ratios -,-,-,■•■ be equal to k. 

X y z 



or. 



Thei 


Q 


a = kx, b 


=^y, 


c = 


^kz, 


'" ) 




,by 


addition. 
















a + 6 + c + •■• 


= k{x 


+ 


y + 


z + 


•••); 






. a + 6 + c 
' X -\- y + z 




= /.- 


_ a 

X 






> 


a : X 


= a. + ?) + 


c+ ••• 


: X 


+ V 


+ 2 


+ •• 



208 GEOMETRY 

VI. .1 giren straight line can he divided iniernalhj in a given 
ratio at one, and only one, point ; and externally at one, and 
only one, point. 



m + ft- 



A?). y._ ^ m-'n X- -■«■-->(?) 



Fig. I. Fig.2. 

Let AB be the p;iven line, and m : n the given ratio, /// ])eing 
gi-eater than n. 

Internal Division, (i) Divide AB (Fig. 1) into m -\- n 
equal parts [Proh. 7] ; and of these parts make AX to con- 
tain m ; then XB must contain n. 

Hence AX : XB = ni : n ; 

that is, AB is divided internally at X in the given ratio, 
(ii) Again, AX : AB = m \ m -\- n. 

Similarly, if P divides AB in the given ratio m : n, 
AP : AB = 7n : ni + n. 

■ AX ^AP, 
" AB AB' 

.: AX = AP. 

Hence P and X coincide ; that is, .Y is the onl^y point 
wliich divides AB internally' in the ratio m : n. 

External Division, (i) Divide AB (Fig. 2) into m — n 
equal parts ; and in AB produced make AX to contain m 
such parts ; then XB must contain n. 

Hence AX : XB = ni : n ; 

that is, AB is divided externally at A' in the given ratio. 

(ii) And it may he siiewn, as above, that X is the only 
point which divides AB externally in the ratio /// : n. 



EXERCISES ON RATIO AND PROPORTION 209 

EXERCISES 

1. Insert the missing terms in the following proportions: 

(i). 3 : 7 = 15 : ( ) ; 
(ii) 2-5: ( )= 10:32; 
(iii) ( ) : ac^ = be: 6c.' 

2. Correct the following statement : 

£ 65 : 78 ft. = £ 25 : 30 ft. 

3. If a straight Line, 9-6" in length, is divided internally in the 
ratio 5 : 7, calculate the lengths of the segments. 

4. If a straight line 4-5 cm. in length is divided externally in the 
ratio 11:8, calculate the lengths of the segments. 

5. AB is a, straight line, 6-4 cm. in length, divided internally at 
X and externally at Y in the ratio 5:3; calculate the lengths of the 
segments, and shew that they satisfy the formula 

AB AX AY 

6. If a sti'aight line, a inches in length, is di\dded internally in 
the ratio m : n, shew that the lengths of the segments are respectively 

— — a inches, — a inches. 

m + ?i m + n 

7. If a straight line, a units in length, is divided externally in the 
ratio m: n, shew that the lengths of the segments are respectively 



'■ — • a units, '■ a units. 

711 — n m — n 

8. It a:h = x: y, and h: c = y : z, prove that a: c = x : z. 

9. If a : 6 = X : y, shew that a -\- h: a = x -\- y: x. 

10. If a, b, c are tlu'ee proportionals, shew that a: c = a"^: b^. 

11. If two straight lines AB, CD are di\'ided internally in the 
same ratio at X and Y respectively, shew that 

(i) AB: XB = CD: YD; 
(ii) AB: AX = CD: CY. 

12. If a, b, c, d are four straight lines such that the rectangle 
contained by a and d is equal to that contained by b and c, prove that 

a:b = c: d. 
p 



210 



GEOMETRY 



PROPORTIONAL DIVISION OF STRAIGHT LINES 
Theorem 46. [Euclid VL 2] 
A straight line drawn parallel to one side of a triangle cuts the 



other two sides, or those sides produced, proportionally. 





In the A ABC, let XY, drawn par' to the side BC, cut 

AB, AC at X and Y, internally in Fig. 1, externally in Fig. 2. 

It is required to prove in both cases that 

AX : XB =^ AY : YC. 

Proof.* Suppose X divides ^Z? in the ratio mm; that 

is, suppose 

AX :XB = m : n ; 

so that, if -4 X is divided into m equal parts, then XB may be 

divided into n such equal parts. 

Through the points of division in AX, XB let parallels be 
drawn to BC. 

Then these parallels divide the segments AY, YC into 
parts which are all equal ; Thcor. 22. 

and of these equal parts A Y contains m, 
and YC contains n ; 
hence A Y : YC = m : n. 

:. AX .XB = AY : YC. q.e.d. 

* Tho proof pivon applies only to fho casf^ in wliioli AX aiul XB 
aro commensurable. The same is true of Theorems 48 and 41). 



PROPORTIONAL DIVISION OF STRAIGHT LINES 211 

Conversely, // a line cuts two sides of a triangle proportion-^ 
ally, it is parallel to the third side. 




Conversely, let XT cut the sides AB, AC proportionalh', so 

that 

AX : XB = AY : YC. 

It is required to prove that XY is parallel to BC. 

Let XP be drawn through A' par' to BC, to meet AC in P. 

Then AP :PC ^ AX : XB ; 

but, by hypothesis, AY : YC = AX : XB. 

Thus AC is cut, internall}' in Fig. 1, and externalh' in Fig. 
2, in the same ratio at P and 5'. 

Hence P coincides with Y, and consequently XP with A" Y. 

Theor. VI, p. 208. 

That is, AFispar' to5C. 

Q.E.D. 

Corollary. If A"}' is parallel to BC , then 

AX -.AB = AY -.AC. 
For, taking Fig. 1, it maj^ be shewn that 
AX : AB = m : m + n ; 
and hence, by Theorem 22, that 

AY -.AC = m : m + n. 
.'. AX \AB = AY .AC. 
Conversely, if AX : AB = AY : AC, 
it may Ije pi'oved as above that AF is par' to BC. 



212 GEOMETRY 

Theorem 47. [Euclid VI. 3 and 4] 

If the vertical angle of a triangle is bisected internally or exter- 
nally, the bisector divides the base iriternally or externally into 
segments which have the same ratio as the other sides of the 
triangle. 

Conversely, if the base is divided internally or externally into 
segments proportional to the other sides of the triangle, the line 
joining the point of section to the vertex bisects the vertical angle 
internally or externally. 




In the A ABC, let AX bisect the Z BAC, internallj^ in 
Fig. 1, and externally in Fig. 2 ; that is, in the latter case, 
let AX bisect the exterior Z B'AC. 
It is required to prove in both cases that 

BX :XC = BA .AC. 
Let CE be drawn through C par' to XA to meet BA (pro- 
duced, if necessar3') at E. In Fig. 1 let B' be taken in AB. 
Proof. Because XA and CE are par', 
.-. , in both Figs., the Z B'AX = the int. opp. Z AEC. 
Also, the Z B'AX = Z XAC 

= the alt. Z ACE. 
.: the Z AEC = the Z ACE. 
.'. AC = AE. 
Again, because XA is par* to CE, a side of the A BCE, 
.: . in both Figs., BX :XC = BAAE ; 

that is, BX :XC = BA :AC. q.e.d. 



INTERNAL AND EXTERNAL DIVISION 213 

Conversely, let BC be divided internally (Fig. 1) or exter- 
nally (Fig. 2) at A', so that BX : AT =^ BA .AC. 

It is required to prove that the Z B'AX = the Z XAC. 

Proof. For, with the same construction as before, 
because A^A is par* to CE, a side of the A BCE, 
.: BX : XC = BA -.AE. 

But, by hypothesis, BX : XC = BA : AC ; 
.: BA : AC = BA : AE ; 

.-. AC = AE. 
.: the Z AEC = the Z ACE 

= the alt. Z XAC. 
And in both Figs., 

the ext. Z B'AX = the int. opp. Z AEC ; 
:. the Z B'AX = the Z XAC. 

Q.E.D. 

Definition 

When a finite straight line is divided internally and exter- 
nally into segments which have the same ratio, it is said to be 
cut harmonically. 

Hence the following Corollary to Theorem 47. 

The base of a triangle is dividedrfiarmonicaUy by the internal 
and external bisectors of the vertical angle; 

for in each case the segments of the base are in the ratio of 
the other sides of the triangle. 



214 GEOMETRY 

EXERCISES ON THEOREM 40 
(N umcrical and Graphical) 

1. On a base AB, 3-5" in length, draw any triangle CAB ; and 
from AB cut off AX 2 1" long. Through X draw AT i)arall(l to 
BC to meet ^C at Y. 

Measure A Y, YC ; and hence compare the ratios 

(i) AK, AX. (H) A^, AC, (Hi) A^^,A£. 

^' XB VC AX AY XB YC 

2. ABC is a triangle, and AT is drawn parallel to BC, cutting 
the other sides at A' and )'. 

(i) If AB = 3-G", AC = 2-4", and .4A' = 21", calculate the 
length of A Y. 

(ii) If AB = 2-0", AC = 1-5", and AY = 0-9", calculate the 
length of BX. 

(iii) If X divides AB in the ratio 8 : 3, and if AC = 8-S cm., find 
AY, YC. 

3. ABC is a triangle, and AT is drawn parallel to BC, cutting 
the other sides produced at A' and 1'. 

(i) U AB = 4-5 cm., AC = 3-5 em., and .4 A' = 7-2 cm., find 
by calculation and measurement the length of A Y. 

(ii) If A' divides A B externally in the ratio 11:4, and \^ AC =4-9 
cm., find the segments of AC. 

(Tfieoretical) 

4. Three parallel straight lines cut any two trausvcrsals propor- 
tionally. 

5. The straight line wliich joins the middle points of the oblique 
sides of a trapezium is parallel to the parallel sides. 

G. Two triangles A BC, DBC stand on the same side of the com- 
mon base BC : and from any point E in BC lines are drawn parallel 
to BA, BD, meeting AC, DC in F and C. Shew that FG is parallel 
to A I). 

7. In a triangle ABC a transversal is drawn to cut the sides 
BC, CA, AB (produced if necessary) at D, E, and F respectively, 
and it makes equal angles with AB and AC; prove tiiat 
BD:CD = BFiCE. 



PROPORTIONAL DIVISION OF STRAIGHT LINES 215 

EXERCISES ON THEOREM 47 
{Numerical and Graphical) 

1. Draw a triangle ABC, making a = 1-5", h = 2-4", and 
c = 3-G". Bisect the angle .4, internally and externally, by lines 
which meet BC and BQ produced at A' and Y. 

Measure BX, XC ; BY, YC; hence evaluate and compare the 

ratios 

BX BY BA 

XC YC' AC' 

2. In the triangle ABC, a = 3-5 cm., b = 5-4 cm., c = 7-2 cm. ; 
and the internal and external bisectors of the Z A meet BC at A" 
and Y. 

Calculate the lengths of the segments into which the base is 
divided at A" and Y respectively ; and verify your results graphically. 

3. Frame constructions, based upon Theorem 47, 
(i) to trisect a straight line of given length ; 

(ii) to divide a given line internally and externally in the ratio 
3:2. 

( Theoretical) 

4. .1 D is a median of the triangle ABC ; and the angles .1 DB, 
ADC are bisected by lines which meet AB, AC at E and F respec- 
tively. Shew that EF is parallel to BC. 

5. A BCD is a quadrilateral; shew that if the bisectors of the 
angles A and C meet on the diagonal BD, the bisectors of the angles 
B and D will meet on AC. 

6. Employ Theorem 47 to shew that in anj- triangle 

(i) the internal bisectors of the three angles are concurrent ; 
(ii) the external bisectors of two angles and the internal bisector 
of the third angle are concurrent. _ 

7. If / is the in-centre of the triangle ABC, and if A I is pro- 
duced to meet BC at A", shew that 

AI: IX = AB + AC: BC. 

8. Given the base of a triangle and the ratio of the other sides, find 
the locus of the vertex. 

9. Construct a triangle, having given the base, the i-atio of the 
other sides, and the vertical angle. 



216 GEOMETRY 

PROPORTIONAL AREAS 
Theorem 48. [Euclid VI. 1] 

The areas of triangles of equal altitude are to one another as 
their bases. * n 





Let ABC, DEF be two triangles of equal altitude, standing 
on the bases BC, EF. 
It is required to prove that 

the A ABC -.the A DEF = BC : EF. 
Proof.* Let the triangles be placed so that the bases BC, 
EF are in the same st. line, and the triangles on the same side 
of the line. 

Join AD ; 

then AD is par' to BF. Def. 2, p. lOL 

Suppose the base BC : the base EF = m : n ; 
so that, if BC is divided into m equal parts, then EF may be 
divided into n such equal parts, in each case by st. lines 
drawn from the vertex to the points of division. 

Then the A ABC, DEF are divided into triangles which 
stand on equal bases, and have the same altitude, and are 
therefore all equal. 

And of these equal A, the A ABC contains ni ; 
and the A DEF contains n. 
.-. the AABC: the A DEF = m : w. 
Hence the A ABC : the A DEF = BC : EF. 

Q.E.D. 
* Sih; footnoU; on p. 1210. 




PROPORTIONAL AREAS 217 

Corollary. The areas of 'parallelograms of equal altitude 
are to one another as their bases. 

For let DB, EG be par"^^ of the same 
altitude, standing on the bases AB, EF. 

Join AC, HF. 
Since the par™7)S = twice the A CAB; 
and the par"" EG = twice the A HEF; 
:. the par"' DB : the par"" EG = 

the ACAB : the A HEF = AB : EF. 

Alternative Proof of Theorem 48 

Let p represent the altitude of each of the A ABC, DEF. 
Then the area of the A ABC = h ■ BC X p ; 
and the area of the A DEF = \ ■ EF X p. 

. A ABC ^ \BCXp ^BC 
" A DEF h-EFX p EF' 

EXERCISES 
{Nuvierical) 

1. Of two triangles 7i,T2 of equal altitude standing on bases of 
6-3" and 5-4" Ti contains 12|^ sq. inches. Find the area of T-2. 

2. The areas of two triangles of equal altitude have the ratio 
24 : 17 ; if the base of the first is 4-2 cm., find the base of the other. 

3. Two triangles lying between the same parallels have bases of 
16-20 m. and 20-70 m. ; find to thg nearest square cm. the area 
of the second triangle, if that of the first is 50- 1204 sq. m. 

4. Two parallelograms whose areas are in the ratio 2-1:3-5 lie 
between the same parallels. If the base of the first is 6-6" in length, 
find the base of the second. 

5. Two triangular fields lie on opposite sides of a common base ; 
and their altitudes with respect to it are 4-20 chains and 3-71 chains. 
If the first field contains 18 acres, find the acreage of the other. 



218 



GEOMETRY 



Theorem 49. [Euclid VI. 33] 

In equal circles, angles, ivhethcr at the centres or circumfer- 
ences, have the same ratio as the arcs on which they stand. 

E F 





Let ABE, CDF be equal circles ; and let the A AGB, 
CHD, and also the A AEB, CFD stand on the arcs AB, CD. 
It is required to prove that 

(i) the Z AGB : the Z CHD = the arc AB : the arc CD; 

(ii) the Z AEB : the Z CFD = the arc AB : the arc CD. 

Proof.* Suppose the arc AB : the arc CD = ??? : n ; 

so that, if the arc AB is divided into m equal parts, then the 

arc CD may be divided into ti such equal parts, in each case 

by radii drawn to the points of division. 

Then the A AGB, CHD, in equal circles, are dividetl into 
angles which stand on equal arcs, and arc therefore all equal. 
And of these equal angles the Z A GB contains m, 
and the Z CHD contains n ; 
.-. the A AGB .the A CHD = m : n. 
Hence the A AGB : the Z CHD = the arc AB: the arc CD. 
And since the Z AEB = one half of the Z AGB; Thcor. 39 
and the Z CFD = one half of the Z CHD; 
.: the A AEB : the Z CFD = the arc AB : the arc CD. 

Q.E.D. 

CoROLL.\RY. Since in equal circles, sectors ichich have equal 
angles are equal [p. 147, E], it may be proved as above that 
the sector AGB : the sector CHD = the arc AB : the arc CD. 
* Sco footnote on p. 210. 




SIMILAR FIGURES 219 

SIMILAR FIGURES 

1. Two rectilineal figures are said to be equiangular to 
one another when the angles of the first, taken in order, arc 
oqusd respectively to those of the second, taken in order. 

2. Rectilineal figures are said to be similar when they are 
equiangular to one another, and also have their correspond- 
ing sides proportional. 

Thus the two quadrilaterals A BCD, 
EFGH are similar if the angles at 
A, B, C, D are respectively equal to 
those at E, F, G, H, and if also rj q 

AB : EF= BC : FG= CD: Gil = DA : HE. 

3. Similar figures are said to be similarly described with 
regard to two sides, when these sides correspond. 

NOTE ON SIMILAR FIGURES 

Similar figures may be desci'ibed as ha^dng the same shape. 

For this, the figures must satisfy two conditions: 
(i) they must have their angles equal each to each, taken in order; 

(ii) their corresponding sides must be proportional. 

In the case of triangles we shall learn that these conditions are not 
independent, for each follows from the other : thus 

(i) if the triangles are equiangular to one another. Theorem ')0 
proves that their corresponding sides arc proportional ; 

(ii) if the triangles have their sides propor- 
tional. Theorem 51 proves that they are 
equiangular to one another. — ~ 

On the other hand, the first diagram 
in the margin shews two figures which 
are equiangular to one another, but 
which clearly have not their sides propor- 
tional ; while the figures in the second 
diagram have their sides proportional, 
but are not equiangular to one another. 




220 GEOMETRY 

SIMILAR TRIAXGLES 
Theorem 50. [Euclid VI. 4] 

7/ two triangles are equiangular to one another, their corre- 
sponding sides are proportional, and the triangles are similar. 

A 





Let the A ABC, DEF have the A A, B, and C respec- 
tively equal to the A D, E, and F. 
It is required to prove that 

AB -.DE = BC : EF = CA : FD. 
Proof. Apply the A DEF to the A ABC, so that E falls 
on B, and EF along BC; 

then since the A E = the A B, ED will fall along BA. 
Let D and F fall at G and H respectively ; so that CBH 
represents the A DEF in its new position. 

Now, by hypothesis, the AD = the A A ; 
that is, the ext. A BGH = the int. opp. A BAG; 
.-. Gil is par' to AC. 
Hence BA : BG = BC : BII; - Thcor. 46, Cor. 

that is, AB : DE = BC : EF. 

Similarly, by api^lying the A DEF to the A ABC, so that 
F falls on C, and FE, FD along CB, CA, it may be shewn 
that 

BC :EF = CA : FD. 

Hence AB : DE = BC : EF = CA : FD, 

and so the triangles are similar (see p. 219). q.e.d. 



SIMILAR TRIANGLES 221 

Theorem 51. [Euclid VI. 5] 

// two triangles have their sides proportional when taken in 
order, the triangles are equiangular to one another, and the 
triangles are similar. 

A D 





B C 

In the A ABC, DEF, let 

AB :DE = BC :EF = CA : FD. 
It is required to prove that the A ABC, DEF are equiangular 
to one another. 

At E in FE make the Z FEG equal to the Z B; 
and at F in EF make the Z EFG equal to the Z C. 

.'. the remaining Z EOF = remaining Z A. 
Proof. Since the A ABC, GEF are equiangular to one 
another, 

.-. AB : GE = BC : EF. Theor. 50. 

But, by hypothesis, AB : DE = BC .EF; 
:. AB :GE = AB : DE. 
:. GE = DE. 
Similarly GF = DF. 
Then in the A GEF, DEF, 

[ GE = DE, GF = DF, 

because <,__,. 

[ and ii/' IS common; 

.'. the triangles are identically equal; Theor. 7. 

. .: the Z DEF = the Z GEF = the Z B; 

and the Z DFE = the Z CF^ = the Z C. 

.'. the remaining Z D = the remaining Z A ; 

that is, the A DEF is equiangular to the A ABC. 

Hence the triangles are similar (see p. 219.) q.e.d. 



AX 


= 1--)"; find 


A Y. 


AY 


= 1-2"; fuul 


AX. 


AY 


= G-G em. ; find 


AC. 


AX 


= 1-4"; find 


XY. 


AX 


= 4-5 cm. ; find 


A H. 


h = 


3G", c = 4-2"; 


and 


0". 


Find the remai 


ning 



222 GEOMETRY 

EXERCISES ON SIMILAR TRIANGLES 

{Numerical and graphical. The results are to be obtained by 
calculation and checked graphically) 

1. In a triangle ABC, XY is drawn parallel to BC, cutting the 
other sides at A' and F : 

(i) If AB = 2-5", AC = 2-0", 
(ii) If AB = 3-5", AC = 2-1", 
(iii) If AB = 4-2 cm., AX = 3-6 cm., 

2. In the figure of the last example : 
(i) li AB = 2-4", BC = 3G", 

(ii) If BC =7-7 em., AT =5-5 em., 

3. In the triangle ABC, a = 3-0", 
QR, drawn parallel to AC, measures 3-0". 
sides of the triangle QBR. 

4. ABC is a triangle in which a = 8 cm., b = 7 cm., and c = 10 
cm. In /IB a point P is talam 4 cm. from A, and P(^ is drawn 
parallel to BC. Find the lengths of PQ and QC. 

5. The sides of a triangular field are 400 yards, 350 yards, and 
300 yards respectively. In a plan of the field the greatest side 
measures 2-4"; find the lengths of the other sides. 

6. XF is drawn parallel to BC, the base of the triangle ABC. 
If AX = 8i ft., A'F = 3^ ft., AY = 6 ft. 2 in., and XB = 4J ft. ; 
calculate the sides of the triangle ABC. 

7. The triangle ABC is right-angled at C ; and from P, a point 
in the hypotenuse, PQ is drawn parallel to AC. 

If AC = li", BC = 3", and PQ = ]"', find BQ, BP, and AP. 

S. In a triangle ABC, AD is the perpendicular from A on BC ; 
and through X, a point in AD, a parallel is drawn to BC, meeting 
the other sides in P, Q. 

If BC = 9 cm., AD = 8 cm., DA = 3 cm. ; find PQ. 

0. In the triangle ABC, a = 2-0 cm., b = 3;') cm., r - 4-5 cm. 
BD and CE are drawn from the ends of the base to the oj)posito 
sides, and they intersect in P. 

If " EP: PC = DP: PB = 2 : f), 

find the lengths of ED, AD, and DC. 



SIMILAR TRIANGLES 223 

EXERCISES OX SIMILAR TRIANGLES 
( Theoretical) 

1. Shew that the straight line which joins the middle points of 
two sides of a triangle is 

(i) parallel to the third side ; (ii) one-half the third side. 

2. In the trapezium A BCD, AB is parallel to DC, and the 
diagonals intersect at : shew that 

0A-: OC = OB-.OD. 
If AB = 2 DC, shew that is a point of trisoction on both 
diagonals. 

3. If three concurrent straight lines are cut by two parallel 
transversals in ^-l, B, C, and P, Q, R respectively; prove that 

AB: BC = PQ:QR. 

4. A BCD is a parallelogram, and from D a straight line is drawn 
to cut AB at E, and CB produced at F. In this figure name thrco 
triangles which are equiangular to one another ; and shew that 

DA : AE = FB: BE = FC: CD. 

5. In the side AC of a triangle ABC any point D is taken : shew 
that if AD, DC, AB, BC are bisected in E, F, G, H respectively 
then EG is equal to HF. 

6. AB and CD are two parallel straight lines; E is the mid- 
point of CD ; AC and BE meet at F, and AE and BD meet at G : 
shew that FG is parallel to AB. 

7. AB h & diameter of a circle, and through ,1 any straight line 
is drawn to cut the circumference in C and the tangent at B and D ; 
shew that 

(i) the A CAB, BAD are equiangular to one another; 
(ii) AC, AB, AD are three propoi-tionals ; 
(iii) the rcct. AC, AD is constant for all positions of AD. 

8. If through any point A^ within^ circle two chords AB, CD 
are drawn, and AC, BD joined; shew that 

(i) the ^ AXC, DXB are equiangular to one another; 
(ii) AX: DX = XC : XB. 

9. If from an external point X a tangent XT and a secant XAB 
are drawn to a circle, and AT, TB joined; shew that 

(i) the A A AT, TXB are equiangular to one another; 
(ii) XAiXT = XT:XB. 



224 GEOMETRY 



Theorem 52. [Euclid VI. 6] 

If two triangles have one angle of the one equal to one angle of 
the other, and the sides about the equal angles proportionals, the 
triangles are similar. 





In the A ABC, DEF, let the Z A = the Z D, 
and let AB : DE = AC : DF. 

It is required to prove that the A ABC, DEF are similar. 

Proof. Apply the A DEF to the A ABC, so that D falls 
on A, and DE along AB; then 

because the Z EDF - the Z BAC, DF must fall along AC. 

Let G and II be the points at which E and F fall respec- 
tively ; so that AGII represents the A DEF in its new 
position. 

Now, by hypothesis, AB : DE = AC .DF; 
that is, AB :AG = AC .AH; 

hence GH is pai-' to BC. Theor. 4G, Cor. 

:. thecxt. ZAGH, namely the LE, = the int. opp. lABC; 
and the ext. I AUG, namely the Z F, = the int. opp. /.ACB. 

Hence Ihe A ABC, DEF are equiangular (o one anotluM', 
hence, the A ABC, DEF are similar. Theor. 50. 

Q.E.D. 



SIMILAR TRIANGLES 225 

♦Theorem 53. [Euclid VI. 7] 

If two triangles have one angle of the one equal to one angle of 
the other, and the sides about another angle in one proportional 
to the corresponding sides of the other, then the third angles arc 
either equal or suppleinentarTj ; and in the former case the tri- 
angles are similar. 

A 




In the A ABC, DEF, let the Z 5 = the Z E ; and let 

AB : DE = AC : DF. 
It is required to prove that 
either the Z. C = the Z F [as in Figs. 1 and 2] ; 
or the A C = the supplement of the Z F [Figs. 1 and 3]. 
Proof, (i) If the Z A = the Z D [Figs. 1 and 2], 

then the I C =^ the Z F; Theor. IG. 

and the A are equiangular, and therefore similar. 

(ii) If the Z A is not equal to the Z D [Figs. 1 and 3], 

let the Z EDF' = the Z A. 
Then the A ABC, DEF' are equiangular to one another ; 

.: AB :DE=AC :DF'. 

But AB :DE = AC^DF; (Hypothesis) 

.: AC : DF' = AC : DF. 

:. DF' = DF. 

:. the Z DFF' = the Z DF'F. 

= the supplement of the Z DF'E 
= the supplement of the Z C. 

Q.E.D. 



226 GEOMETRY 

EXERCISES ON SIMILAR TRIANGLES 
( Theoretical) 

1. In a triangle ABC, prove that any straight line parallel to 
the base BC and intercepted by the other two sides is bisected by 
the median drawn from the vertex A. 

2. Two triangles ABC, A'B'C are equiangular to one another; 
if p, p' denote the perpendiculars from A, A' to the opp. sides 

R, li' circum-radii ; 

r, r' in-radii ; 

prove that each of the ratios ^ , — , -^ is equal to the ratio of any 
pair of corresponding sides. P >' ^ 

3. Prove that the radius of the circle which passes through the 
mid-points of the sides of a triangle is half the circum-radius. 

4. If two straight lines AB, CD intersect at A', so that 

XA-.XC = XD.XB; 

(i) shew by Theorem 52 that the A AXD, CXB are similar; 

(ii) hence prove the points A, D, B, C concyclic. 

o. A, B, C are three coUinear points, and from B and C two 

parallel lines BP, CQ are drawn in the same sense, so that 

PB-.QC = AB: AC; 

shew by Theorem ")2 that the points A, P, Q are collincar. 

G. If in tAvo triangles ABC, A'B'C, the Z B = the Z B', and 

r h 

— = — ; what conclusion mav be drawn ? 

c' b' 

Shew by diagrams how this conclusion is affeetcd, if it is also 

given that 

(i) c is less than b, 

(ii) c is equal to b, 

(iii) c is greater than h. 

7. .1 BCD is a parallelogram ; P and Q arc points in a straight 
line parallel to AB; PA and QB meet at R, and PD and QC meet 
at S : shew that RS is parallel to A D. 

8. In a triangle ,4 BC the bisector of the ^'ertical angle A meets 

the base at D and the circumference of the circum-cirde at E; if 

EC is joined, shew that the triangles BAD, EAC are similar; and 

hence prove that 

ABAC = AE-AD. 



SIMILAR TRIANGLES 227 

Theorem 54. [Euclid VI. 8] 

In a right-angled triangle, if a perpendicular is drawn from 
the right angle to the hypotenuse, the triangles on each side of it 
are similar to the whole triangle and to one another. 

A 




Let BAC be a triangle right-angled at A, and let AD be 
drawn perp. to BC. 

It is required to prove that the A BDA, ADC are similar to 
the A BAC and to one another. 

In the A BDA, BAC 
the Z BDA = the Z BAC, being rt. angles, 
and the Z B is common to both; 
.'. the remaining ZB.4Z) = the remaining ZBCA ; Theor.lQ. 
hence the A BDA is equiangular to the A BAC; 
.'. their corresponding sides are proportional; 
.". the A BDA, BAC are similar. 
Similarly the A ADC, BAC may be proved similar. 
Hence the A BDA^, ADC, being equiangular to the A BAC, 
are equiangular and hence similar to each other. q.e.d. 

Corollary, (i) Because the is DBA, DAC are similar, 
.-. DB: DA = DA: DC; 
that is, DA is a mean proportionaHjetween DB and DC; 
and hence DA- = DB ■ DC. 

(ii) Because the A BCA, BAD are similar, 
.-. BC: BA = BA: BD; 
hence BA^- = BCBD. 

(iii) Because the ^ CBA, CAD are similar, 
.-. CB: CA = CA: CD; 
hence CA^ = CB-CD. 



228 GEOMETRY 

EXERCISES 

(Miscellaneous Examples on Theorems 50-54) 

1. ABC is an equilateral triangle of which each side = a. In 
BC, produced both ways, two points P and Q are taken, such that 
BP = CQ = a, and AP, AQ arc joined. Shew that 

(i) PQ: PA = PA: PB. 
(ii) PA'- = 3a2. 

2. A BC is a triangle, right-angled at A , and AD is drawn per- 
pendicular to BC : i{ AB, AC measure respectively 4" and 3", shew 
that the segments of the hj'potenuse are 3-2" and 1-8". 

3. A BC is a triangle right-angled at A , and a perpendicular A D 
is drawn to the hypotenuse BC ; shew (i) by Theorem 25, (ii) by 
Theorem 54 that 

BC-AD = AB-AC. 

4. ABC is a triangle right-angled at A, and AC is drawn per- 
pendicular to the hypotenuse, also -C'A' is drawn parallel to CA. 
If AC = 15 cm., and AB = 20 cm., shew that AC = 12 cm., and 
CA' = 9-6 cm. 

, 5. At the extremities of a diameter of a circle, whose centre is 
C and radius r, tangents are drawn : these are cut in Q and R by 
any third tangent whose point of contact is P. Shew that 
(i) QR subtends a right angle at C ; 
(ii) PQPR = r2. 

6. Two circles of radii r and r' respectively have external con- 
tact at A, and a common tangent touches them at P and Q. Shew 

that 

(i) PQ subtends a right angle at A ; [Ex. 9. p. 182.] 

(ii) PQ"" = 4 rr'. 
[Produce PA, QA to meet the c'irfMunf(>rences at X and }', and 
prove the triangles PAY, X AQ riglit-anglrd and similar.) 

7. Two circles touch one another externally at .4, and a com- 
mon tangent PQ is produced to meet the line of centres at .S'. Shew 
that, if PA, AQ are joined, 

(i) the triangles SAP, SQA are similar; 
(ii) SA^ = SP- SQ. 

8. Two circles intersect at A and B ; and at .4 tangents are 
drawn, one to each circle, to meet the circumferences at C and D: 
shew that if BC, BD are joined, then BC : BA = BA : BD. 



AREAS OF SliMlLAU TRIANGLES 



229 



Theorem 55. [Euclid M. 10] 

The areas of similar triangles are proportiotial to the squares 
on correspondimj sides. 
A 





Let ABC, DEF be i^iiiiilar triangles, in which BC and EF 

are corresponding sides. 

It is required to -prove that 

the A ABC : the A DEF = BC~ : EF\ 

Let AG and DH be drawn pcrp. to BC, EF respectively; 

and denote these perp^. by p and p' . 

Proof. The A ABC = \BC ■ p ; the A DEF = \EF- ;/. 

. ^ABC _ BC p ... 

"a DEF EFp' ^^^' 

But since the /LB — the Z E, from the similar A 

ABC, DEF, 

and the Z. G = the Z H, being right angles ; 

.'. the A ABG, DEH are equiangular to one another, 

Theor. 16. 

• P -A^ 
" p' DE 

BC 



Theor. 50. 



= ^, from the sunilar A ABC, DEF. 
Er 

Substituting for ^ in (i), 
V 

A ABC ^ BC BC ^ BC^ . 
EF^' 



or, 



A DEF EF ■ EF 
the A ABC : the A DEF = BC- : EF'-. q.e.d. 



230 GEOMETRY 

EXERCISES ON THE AREAS OF SIMILAR TRIANGLES 

{Numerical and Graphical) 

1. In any triangle ABC, the sides AB, AC are cut by a lino A'}' 
drawn parallel to BC. If ylA is one-third of AB, what j)art is the 
triangle AXY of the triangle ABC? 

2. Two corresponding sides of similar triangles are 3 ft. G in. 
and 2 ft. 4 in. respectively. If the area of the greater triangle is 
45 sq. ft., find that of the smaller. 

3. The area of the triangle A BC is 2.5-6 sq. cm., and A' Y, drawn 
parallel to BC, cuts AB in the ratio 5:3. Find the area of llu^ 
triangle ,1A"F. 

4. Two .similar triangles ha^^e areas of 392 sq. cm. and 20() sc]. 
cm. respectively ; find the ratio of any pair of corresponding sides. 

5. A BC and A YZ are two similar triangles whose areas are 
respectively 32 sq. in. and 60-5 sq. in. If XY = 7-7", find the 
length of the corresponding side AB. 

6. Shew how to draw a straight line A'l' parallel to BC the liase 
of a triangle ABC, so that the ftrea of the triangle AXY may bo 
nine-sixteenths of that of the triangle ABC. 

( Theoretical) 

7. ABC is a triangle, right-angled at .1, and AD is drawn jxt- 
pendicular to BC ; shew that 

A BAD: A ACD = BA"-: ACK 

8. A trapezium A BC D has its sides AB, CD i);irallcl, and lis 
diagonals intersect at O. }f A B is double of CD, find the ratio of 
the triangle AOB to the triangle COD. 

9. If two triangles have one angle of one equal lo ono angle of 
the other, their areas are proportional to the rectangles contained 
by the sides about the equal angl(>s. 

10. Prove that the areas of similar triangles have the same ratio 
as the squares of 

(i) corresponding altitudes; 

(ii) corresponding medians; 

(iii) the radii of their in-circles; 

(iv) the radii of their circum-circlcs. 



CHORDS OF CIRCLES 231 

RECTANGLES IN CONNECTION WITH CIRCLES 
Theorem 56. [Euclid III. 35 and 36] 

If any two chords of a circle cut one another internally or 
externally, the rectangle contained hy the segments of one is 
equal to the rectangle contained by the segments of the otJur. 





Fiff. X, Fig. 2. 

In the0 ABC, let the chords AB, CD cut one another at 
A', internally in Fig. 1, and externally in Fig. 2. 
It is required to prove in both cases that 

the recL XA, XB = the red. AT, XD. 
Join AD, BC. 

Proof. In the A AXD, CXB, 

the Z AXD = the Z CXB, being opp. vert. A in Fig. 1, 
and the same angle in Fig. 2; 

and the Z A = the Z C, being A at the O^'^ standing on the 
same arc BD; 

.: the remaining angles^ are equal; Theor. 16. 

hence the A AXD, CXB are equiangular, 

.XA^XD, 

" XC XB' 

:. XAXB = XC-XD; 

that is, the rect. XA, XB = the rent. AT, XD. q.e.d. 



232 



GEOMETRY 



Corollary. If from an external point a secant and a tan- 
gent are drawn to a circle, the rectangle contained by the whole 
secant and the part of it outside the circle is equal to the square 
on the tangent. 




Let XBA be a secant, and XT a tangent drawn to the 
O AB Thorn the point X. 
It is required to prove that XA ■ XB = X T-. 



Proof. 
Then 



because 



io\n AT, BT. 

•the A X AT = the Z XTB, Theor. 45. 
the Z TXA is common, 

. the third angles are equal, Theor. 16. 

the A XAT, XBT arc similar. Theor. 50. 

' ' XT XB' 

.'. rect. XA, XB = sq. on XT. q.e.d. 



RECTANGLES IN CONNECTION WITH CIRCLES 233 

EXERCISES ON THEOREM 56 
( Theoretical) 

1. ABC is a triangle right-angled at C ; and from C a perpen- 
dicular CD is drawn to the hypotenuse; shew that 

ADDB = CD'-. 

2. If two circles intersect, and through any point X in their 
common chord two chords AB, CD are drawn, one in each circle, 

shew that 

AX- XB = CX- XD. 

3. Deduce from Theorem 56 that the tangents drawn to a circle 
from any external point are equal. 

4. If two circles intersect, tangents drawn to them from any 
point in their common chord produced are equal. 

5. If a common tangent PQ is drawn to two circles which cut 
at A and B, shew that AB produced bisects PQ. 

6. If two straight lines AB, CD intersect at A' so that ^A'- XB 
= CX ■ XD, deduce from Theorem 56 (by reductio ad ahsurdum) that 

the points A, B, C, D are concyclic. 

7. In the triangle A BC, perpendiculars A P, BQ are drawn from 
A and B to the opposite sides, and intersect at ; shew that 

AO-OP = BO- OQ. 

8. ABC is a triangle right-angled at C, and from C a perpen- 
dicular CD is drawn to the hypotenuse; shew that 

AB-AD = AC: 

9. Through A, a point of intersection of two circles, two 
straight lines CAE, DAF are drawn, each passing through a centre 
and terminated by the circumferences ; shew that 

CA-AE = DA-AF. 

10. If from any external point-P two tangents are dra\\Ti to a 
given circle whose centre is and radius r; and if OP meets the 
chord of contact at Q, shew that 

OP-OQ = r2. 

11. AB {?, K fixed diameter of a eii-cle, and CD is perpendicular 
to AB {or AB produced) ; if any straight line is drawn from A to 
cut CD at P and the circle at Q, shew that 

AF- .\Q = constant. 



234 GEOMETRY 

EXERCISES ON THEOREM 56 

(Miscellaneous) 

1. The chord of an arc of a cu-ele = 2c, the height of the arc = h, 
the radius = r. Shew by Theorem 56 that 

h{2r - h) = c\ 
Hence find the diameter of a circle in which a chord 24" long cuts 
off a segment 8" in height. 

2. The radius of a circular arch is 25 feet, and its height is 18 
feet ; find the span of the arch. 

If the height is reduced by 8 feet, the radius remaining the same, 
by how much will the span be reduced? 

Check your calculated results graphically by a diagi'am in which 
1" represents 10 feet. 

3. Employ the equation h{2r — h) = c^ to find the height of an 
arc whose chord is 16 cm., and radius 17 em. 

Explain the double result geometrically. 

4. If d denotes the shortest distance from an external point to 
a circle, and t the length of the tangent from the same point, shew 

by Theorem 56 that 

did + 2r) = t\ 

Hence find the diameter of the circle when d = 1-2", and 
I = 2-4"; and verify your result graphicallj'. 

5. If the horizon visible to an observer on a cliff 330 feet above 
the sea-level is 22J miles distant, find roughly the diameter of the 
earth. 

Hence find the approximate distance at which a bright light 
raised 66 feet above the sea is visible at the sea-levol. 

6. If h is the height of an arc of radius r, and h the chord oi' lialf 
the are, prove that 6" = 2rh. 

7. A semi-circle is described on yl /? as diameter, and any two 
chords AC, BD are drawn intersecting at P; shew that 

AB^ = ACAP + BDBP. 

8. Two circles intersect at B and C, and the two direct common 
tangents AE and DF are drawn; if the common chord is produced 
to meet the tangents at C, and //, shew tliat 

(111- = AK' + B('\ 



FOURTH PIIOPUKTIONALS 



235 




PROBLEMS 
Problem 33 
To find the fourth proportional to three given straight lines. 

Let A, B, C be the three given 
st. lines, to which the fourth pro- 
portional is required. 

ACB 

Construction. Draw two st. lines DL, DK of indefinite 
length, containing any angle. 

From DL cut off DG equal to A, and GE equal to B ; 

and from DK cut off DH equal to C. 

Join GH. Through E draw EF par" to GH. 

Then HF is the fourth proportional to ^ , B, C. 

Proof. Because GH is par' to EF, a side of the A DEF ; 
.-. DG : GE = DH : HF. 
That is, A : B = C : HF. 
Then HF is the fourth proportional to .4, B, C. 



Problem 34 
To find the third proportional to two given straight lines. 

Let A, B he the two lines to 

which the third proportional is 

required. . 

AB D G E •!. 

This problem is that special case of Problem 36 in which 
C = B. (See 5, p. 205.) The solution given alx)ve appHes 
to it. 




236 



GEOMETRY 



Problem 35 

To divide a given straight line internally and externally in a 
given ratio. 



M N A 




X B 



Let AB he the st. line to be divided internally and exter- 
nally in the ratio M : A'^. 

Construction. At A make any angle BAH with AB. 

From AH cut off AP equal to M. 

From PH and PA cut off PC and PC, each equal to N. 

Join BC, BC. 
Through P draw PX par' to BC, and PY par' to BC. 
Then AB is divided internally at X, and externally at Y 
in the ratio M : N. 

Proof, (i) Because PX is par' to BC, a side of the A ABC, 
.'. AX : XB = AP :PC = M :N. 
(ii) Because P 7 is par' to BC, a side of the A ABC, 
.'. A Y : YB = AP : PC = M : N. 

Corollary. By a similar prr)c(>ss a 
st. line AB may be divided internally 
into segments proportional to three lines. 

Construction. Draw AH, and from it 
cut off AP, PQ, QR equal respectively to 
L, M, N. Join RB ; and'through P and 
QdrawPX, ()y par' to BR. 

Then evidently 

AX : L ^ XY .M = YB 




Y B 



A^. 



MEAN PROPORTIONALS 



237 



Problem 36 

To find the mean proportional hetween two given straight 
lines. 




Let AB, AC be the two given st. lines. 

Construction. Place AB, AC in a straight line, and in 
opposite senses ; and on BC describe the semi-circle BDC. 
From A draw AD at rt. angles to BC, to cut the O" at D. 
Then AD is the mean proportional between AB and AC. 

Proof. Join BD, DC. 

Now the Z BDC, being in a semi-circle, is a rt. angle. 
And in the right-angled A BDC, DA is perp. to BC, 

.'. the A ABD, ADC are similar ; Theor. 54. 
.-. AB -.AD = AD :AC ; 
that is, AD is the mean proportional between AB and AC. 



Note. If the given lines AB, AC arc 
placed in the same sense, the mean propor- 
tional between tliem may be cut off from A B 
by the following useful construction. 



On AB draw a semi-circle; and from C draw CD perp. to AB 
to cut the O"^ at D. From AB cut off AX equal to AD. 

Then AX is the mean pi-oportional between AB and AC. 

For the ^ ABD, ADC are similar, Theor. 54. 

.-. AB: AD = AD: AC; 
that is, AB: AX = AX: AC. 




238 GEOMETRY 

GRAPHICAL EVALUATION OF A QUADRATIC SURD 

Example. Find the approximate value of (i) Vo, (ii) v'21. 

(i) Vs = V5 X 1. Hence take AB, AC rcspoctively to repre- 
sent 5 and 1 in terms of any convenient unit, and find A D, the mean 
proportional between them. 

Then A D- = AB-AC III, p. 200. 

= 5X1 =5. 
.-. AD = VS. 

By measuring AD, the value of V5 is roughly found to l)o 2-24. 

(ii) V2I = V7 X 3. Here take AB, AC equal to 7 cm. and 3 
cm. respectively, and proceed as before. 

Note. Factors should be chosen so as to give convenient lengths 
for AB, AC. 

e.g. \/23 = V2-3 X 10; vU = V2^'x^. 

EXERCISES 

1. Find graphically, testing your results by arithmetic : 

(i) The 4th proportional to 2-4", 1-5", 1-6". 
(ii) The 3rd proportional to 2-5" and 1-5". 
(iii) The mean proportional between 7-2 cm. and 50 cm. 

2. Di\'ide a line, 20" in length, internally and externally in the 
ratio 7:3; and in each case measure and calculate the segments. 

3. Obtain graphically the unknown term in the following state- 
ments of proportion ; and check your result by arithmetic : 

(i) 1-25: X = 10: 1-6. (Take 1" as the unit of length.] 
(ii) x: 4-2= 4-2: 6-3. (Take 1 cm. as the unit of length.] 
(iii) x:16= 25 : .r. [Ix-t 1" represent 10.] 

4. Divide a line, 7-2 cm. in length, into three parts proportional 
to the numbers 2, 3, 4. Measure and calculate these parts. 

5. Divide a line, 3-9" in length, into three parts, so that the 
second = | of the first, and the third = J of the second. 

G. On a side of 1-5" draw a rectangle equal in area to a square 
on a side of 2". Measure the other side of the rectangle. 
7. Find graphically the approximate A\alues of 
(i) VS; (iij VlO; (iii) V^i. 



EXERCISES IN PROPORTION 239 

8. Determine geometrically the approximate values of tlif f(jl- 
lowing expressions, verifying each drawing arithmetically : 

(i) 3-5 X 2-4 (ji) 6:M, (iii) 2-71 X 1-2(3. 
^' 2-8 ^ 2-13 1-51 

9. Draw a triangle .4 BC from each of the following sets of data, 
and in each case calculate and measure the lengths of the sides : 

(i) The perimeter = 4-8"; and a: 3 = ?>: 4 = c: 5. 

(ii) The perimeter = 11-1 em. ; and a — lb, b = ^ c. 
(iii) The perimeter = 11-8 em. ; and .4 : 1 = /? : 2 = T: .3. 
(iv) a = 40", .4 = 90°; and 6: c = 5:3. 

10. A field is represented in a plan by a triangle A BC, in whieh 
n = 8 em., 6 = .5Gem.,c = G-4f'm. If the gi-eatest side of the field 
is 200 metres, find the lengths of the other sides. 

A fence, run across the field, is represented in the plan by a line 
PQ parallel to BC drawn from a point P in .1 B distant 40 cm. from 
A . Find the length of the fence. 

11. A man 6 feet in height, standing 1.5 feet from a lamp-post, 
observes that his shadow cast by the light is 5 feet in length ; how 
high is the light, and how long would his shadow be if he were to 
approach 8 feet nearer to the post? 

12. To find the width of a canal a rod is fixed vertically on the 
bank so as to shew 4^ feet of its length. The observer, whose eye 
is 5 ft. 8 in. above the gi'ound, retires at right angles from the canal 
until he sees the top of the rod in a line with the further bank. If 
his distance from the canal is now 20 feet, what is its width ? 

13. A man, wishing to ascertain the height of a tower, fixes a 
staff vertically in the ground at a distance of 27 ft. from the tower. 
Then, retiring 3 ft. farther from the-tower, he sees the top of the 
staff in line with the top of the tower. If the observer's eye and 
the top of the staff are respectively 5 ft. 4 in. and 12 ft. above the 
ground, find the height of the tower. 

14. A person due S. of a Ughthouse observes that his shadow 
cast by the Hght at the top is 24 feet long. On walking 100 yards 
due E. he finds his shadow to be 30 feet long. Supposing him to be 
G feet high, find the height of the light from the ground. 



240 GEOMETRY 

SIMILAR POLYGONS 
Theorem 57 

Similar polygons can he divided into the same number of 
similar triangles; and the lines joining corresponding vertices 
in each figure are proportional. 





Let ABODE, FGHKL be similar polygons, the vertex A 
corresponding to the vertex F, B to G, and so on. Let AC, 
AD be joined, and also FH, FK. 

It is required to prove that 

(i) the A ABC, FGH are similar ; as also the A ACD, 
FHK, and the A ADE, FKL. 

(ii) AB :FG = AC : FH = AD : FK. 

Proof. (i) Since the polygons arc similar,. 

the Z ABC = the Z FGH, and AB : FG = BC : GH ; 

.: the A ABC, FGH are similar. Thcor. 52. 

.-. the Z BCA = the Z GHF ; 
Also the Z BCD = the Z GHK ; 
/. the £ ACD = the Z FHK. 

Also AC : FH = BC : GH (the A being similar) 

= CD : HK (the polygons being similar). 
/. the A ACD, FHK are similar. Thcor. 52. 
In the same way the A ADE, FKL are similar. 



SIMILAR POLYGONS 



241 



(ii) And AB : FG = AC :FH, from the similar A ABC, 
FGH ; = AD : FK, from the similar A CAD, 

HFK. Q.E.D. 

Note. In Theorem 57 the polygons have been divided into simi- 
lar triangles by lines drawn from a pair of corresponding vertices. 
Other ways in which this sub- 
division may be made are: 

(i) By lines drawn from a 
fair of corresponding points on 
the perimeters oj the figures, brit 
■not vertices. 

(ii) By lirtes drawn from a 
pair of corresponding points 
within the polygons. 

The proofs of the proposi- 
tion for these cases are Itft as 
an exercise for the student. 

It is well to notice also the 
following case in which the 
subdivision is made. 

(iii) By lines drawn from a pair of corresponding points outside the 
polygons. , 

In this case let the student prove that 
the corresponding triangles are similar 
and note that these triangles are not parts 
of the polygons but that the polygon 
ABODE = the sum of the ^ OAB, OBC, 
OCD, ODE diminished by the A OAE; 
and similarly for the polygon FGHKL. 





242 



GEOMETRY 



Problem 37. [First Method.] 

On a side of given length to draw a figure similar to a given 
rectilineal figure. 




M 



Let ABODE be the given figure, and LM the length of the 
given side ; and suppose that this side is to correspond to 
AB. 

Construction. From AB cut off AB' equal to LM. 
Join AC, AD. 

From B' draw B'C par" to BC, to cut AC at C. 
From C draw CD' pai-" to CD, to cut AD at D'. 
From D' draw D'E' pai-" to DE, to cut EA at £". 
Then A B'C D'E' is the required figure. 

Outline of Proof, (i) By construction the figure AB'CD'E' 
is equiangular to the figure ABODE. 

(ii) From the three pairs of similar triangles it may be 
shewn that 

AB' ^ BT^ ^ CW ^ D'E' _ E'A . 
.AB BC CD DE EA ' 

that is, corresponding sides of (he polygons are proportional. 
Accordingly the figure AB'CD'E' described on a liiiee(iual 
to LM is similar to ABODE. 



SIMILAR FIGURES 



243 



Theorem 58 

Any two similar rectilineal figures may be so placed that the 
lines joining corresponding vortices are concurrent. 




C D 

Fig. I. 

Let ABCD, A'B'C'D' be similar figures. 
Then since the /. B' = the Z B, the figures can be so 
placed that A'B', B'C are respectively par| to AB, BC. It 
follows, since the figures are equiangular to one another, that 
C'D'is par' to CD, and D'A' paH to DA. 

It is required to prove that when corresponding sides of the 
figures are parallel, AA' , BB' , CC , DD' are concurrent. 
Join A A' ; divide it externally at S in the ratio AB : A'B'. 
Join SB and SB' ; it will be shewn that >S5 and SB' are in 
one straight line. 

Proof. In the A SAB, SA'B', since AB and A'B' are par', 
.-. the Z SAB = the Z SA'B' ; 
and, by construction, SA : SA' = AB -.A'B' ; 
.-. the A SAB, SA'B' are similar; Theor. 52. 

.'. the Z ASB = the ZA'SB'. 
Hence SB, SB' are in the same st. line ; 
that is, BB' passes through the fixed point S. 
Similarl}^ CC and DD' may be shown to pass through S. 
That is, AA', BB', CC, DD' are concurrent, q.e.d. 

XoTE. Observe that the joining lines A A', BB', CC, DD' are 
all divided externally at .S' in the ratio of any pair of corresponding 
sides of the given figiires. S is called the centre of similarity. 



244 



GEOMETRY 



NoTK. In placing tlio giv(>n figures so that A'B', B'C are rc- 
spcftivcly parallel to A />, BC, two cases arise : 

(i) A'B' and AB may have the same sense, as in Figs. 1 and 2; 
(ii) A' B' and .4 B may have opposite senses, as in the Fig. below. 
A 




In the latter ease it follows also that CD' is par' to CD, and D'A' 
par' to DA, and it may be proved that A A', BB', CC, DD' are eon- 
current ; but here S dividers A A' internally in the ratio AB: A'B'. 

Problem 37. [Second Method.] 
On a given side to draw a figure similar to a given figure. 

A 




C D 

Let A BCD be the given figure, and A'B' the given side ; 
and let A'B' correspond to AB. 

Construction. Place yl7i' par' to ^B ; and join AA', BB' 

by lines meeting at ;S. 

Join SC, SD. 
Through B' draw B'C par' to BC, to meet SC at C ; 
through C draw CD' par' to CD, to meet SD at D'. 
Join A'D'. 
Then A'B' CD' is the required figure. 
The student should pi'ove (i) that A'B'CD' is equiangular 
to A BCD, (ii) that corresponding sides of these figures are 
proportional. Tlie proof is the converse of Tiieorem 58. 



I 



SIMILAR FIGURES 245 

EXERCISES ON SIMILAR FIGURES 

{Numerical and Graphical) 

1. On a base AB, 6-5 cm. in length, draw a quadrilateral 
A BCD from the following data : 

£ A =^ 80°, Z JS = 70°, AD =4-4 em., BC = 3-2 cm. 
Taking any convenient point as centre of similarity, make 
(i) A reduced copy of A BCD, such that the ratio of each side to 

the corresponding side of A BCD is 3: 4. 

(ii) An enlarged copy of ABCD, such that the ratio of each side 

to the corresponding side of ABCD is 5:4. 

2. In a semi-circle drawn on a given diameter AB, inscribe a 
square, so that two vertices may be on the arc, and two on A B. 

If .1 B = 2?-, and the side of the inscribed square = a, shew that 
5a- = 4r'. 

3. Draw a sector of a circle of radius 2-4", the central angle 
being 60° ; and inscribe a square in it. 

If the radius of the sector = r, and the side of the square = a, 
calculate from measurements the ratio a : r. 

4.- In a sector of which the radius =5 em., and the central angle 
= 45°, inscribe a rectangle with its sides in the ratio 2:1. 

Prove that two such rectangles can be drawn, and compare by 
measurement their greater sides. 

5. Draw a triangle ABC, making a = 8 cm., 6=7 em., and 
c = G cm. 

Working from the vertex ^1 as centre of similarity, inscribe a 
square in the triangle, so that two of its angular points may be in 
the base BC, and the other two in AB, AC. 

G. Draw a triangle ABC, making a = 2G", B = 110°, C = 35°. 

In the triangle ABC inscribe an equilateral triangle, having 

(i) one side parallel to BC; ^ 

(ii) one side parallel to any given straight line. 

7. In a given triangle ABC inscribe a triangle similar to a given 
triangle DEF. 

In how many ways may this be done? 

8. Draw a regular he.xagon ABC DEF on a side of 1 -2", and in 
it inscribe a square having "two sides parallel to AB and DE, and 
its vertices on the remaining sides of the hexagon. 



246 GEOMETRY 

Theorem 59. [Euclid VI. 20] 

The areas of similar polygons are proportional to the squares 
on corresponding sides. 
D 

K 

H 





Let ABODE, FGHKL ho similar polygons, and let AB 
FG be corresponding sides. 
It is required to prove that 
the polygon ABODE : the polygon FCHKL = AB- : FCK 
Join AO, AD, FII, FK. 

Proof. Then the A ABC, FGH are similar ; Thcor. o7. 
also the A AOD, FHK are similar ; 
and the A ADE, FKL are similar. 
.-. the AABO: the A FGH = AO^- : FH- Thcor. na. 

= the AAOD: the A FIIK. 
Similarly', 
the A AOB : the A FHK = AD""- : FK- 

= the A ADE :lhe A FK-L. 

Hence A ABC ^ A AOD ^ A ADE 

A FGH A FHK A FKL 
And in this series of equal ratios, the sum of the ante- 
cedents is to the sum of the consequents as each antecedent 
is to its consequent ; Thcor. \, p. 207. 

.-. the fig. ABODE : the fig. FGHKL 

= the AABO : the A FGH 
= AB^ : FG-. 

Q.E.D. 



AREAS OF SIMILAR POLYGONS 247 

Corollary 1. Let a, h, c represent three lines in pro- 
portion, so that - = - ; and consequently h^ = ac. 





Q 



K 





I 



d c 

Now suppose similar figures P and Q to be drawn on a and 
b as corresponding sides, 
then Fig. P ^ ^ ^ i^ = ^ . 

Fig. Q b^ ac c 

Hence if three straight lines are proporfiouals, and awj 
similar figures are drawn on the first and second a^ correspond- 
ing sides, then 
the fig. on the first : the fig. on the second = the first 

Corollary 2. Let 

AB : CD = EF : GH ; 

A 

and let similar figures KAB,LCD 

be similarly described on AB, CD, M 
and also let similar figures MF, /^**v^ N 

XH be similarly described on / \ / \ 

EF, GH. E F G H 

^, . AB EF . AB- EF' 

Thensmce ^ = ^,; -. e^. = ^.- 

But the fig. KAB : the fig. LCD = AB- : CD- ; Theor. 59. 
and the fig. MF : the fig. XH = EF- : GH-. 
:. the fig. KAB : the fig. LCD = the fig. MF : the fig. XH. 

Hence if four straight lines are proportional, and a pair of 
similar rectilineal figures are similarly described on the first 
and second, and also a pair on the third and fourth, these figures 
o,re proportional. 



24S geomp:try 

EXERCISES 

1. Similar figures are described on the side and diagonal of a 
square ; prove that the ratio of their areas is 1 : 2. 

2. Similar figures are described on the side and altitude of an 
equilateral triangle ; prove that the ratio of their areas is 4 : 3. 

3. The area of a regular pentagon on a side of 2-5" is approxi- 
mately lOj sq. in. ; find the area of a similar figiu'e on a side of 30". 

4. The length of a rectangular area is 10-8 metres, and the 
ratio of the length to the breadth is 12 : 5 ; find the length and 
breadth of a similar rectangle containing one-ninth of the area. 

5. In the plan of a certain field, 1" represents G6 yards ; if the 
area of the plan is found to be 100 sq. in., find the area of the field 
in acres. 

Explain why in this example the shape of the field is immaterial. 

6. An estate is represented on a plan by a quadrilateral A BC D 
drawn to the scale of 25" to the mile. If AC = 20", and the off- 
sets from AC to B and D measure 24" and 26" respectively, find 
the acreage of the estate. 

7. A field of 1-89 hectares is represented on a plan by a triangle 
v/hose sides measure 13 cm., 14 cm., and 15 cm. On what scale is 
the plan drawn? 

8. A regular hexagon is drawn on a side of a em. and a second 
hexagon is inscribed in it liy joining the middl(> points of the sides 
in order. In like manncT a third hexagon is insc-ribed in the second ; 
and so on. Find the ratio of the first hexagon to the fifth. 

9. Compare the area of any regular hcwagon with the areas of (lie 
regular hexagons described on two unequal diagonals of the original 
one. 

10. Compare the areas of the regular inscribed and the regular 
circumscribed hexagons of any circle. 

11. Shew that the areas of two similar cyclic figures are propor- 
tional to the squares of the diameters of their circum-cirdes. 

12. Two similar polygons which are equal in area are equal in 
all respects. 



SIMILAR FIGURES 249 

Theorem GO. [Euclitr VI. 31] 

In a right-angled triangle, any rectilineal figure described on 
the hypotenuse is equal to the sum of the two similar and simi- 
larly described figures on the sides containing the right a?igle. 




Let ABC be a right-angled triangle of which BC is the 
hypotenuse ; and let P, Q, R be similar and similarly de- 
scribed figures on BC, CA, AB respectively. 

It is required to prove that 

the fig. R + the fig. Q = the fig. P. 
Proof. Since AB and BC are corresponding sides of the 
similar figs. R and P, 

. fig. R AB^- 



I 



fig. P BC'- 
In like manner, fig. Q _ AC- 

fig. P~ BC'- ' ' ' 
Adding the equal ratios on each side in (i) and (ii) 
fig. R + fig. Q _ AB'- + AC- 



(i) Theor. 59. 
(ii) 



fig. P BC- 

But AB-' -\- AC- = BC- ; Theor. 29. 

/. the fig. R + the fig. Q-=^ the fig. P. q.e.d. 

Corollary. The area of a circle drawn on the hypotenuse 
of a right-angled triangle as diameter is equal to the su7n of the 
circles similarly drawn on the other sides. 

For the areas of circles are proportional to the squares on 
their diameters. [Page 199.] 



250 GEOMETUY 

EXERCISES 

(Miscellaneous) 

1. In a triangle ABC, right-angled at A, AD is drawn perpen- 
dicular to the hypotenuse. Shew that 

(i) BA'- = BC-BD; (ii) CA^ - CB-CD. 
Hence deduce Theorem 29, namely, 

BC^ = BA^ + ACK 

2. In the diagram of Theorem 60, draw AD perpendicular to 
BC ; hence prove that, if the fig. P = the A A BC, then 

(i) the fig. Q = the A ADC; (ii) the fig. R = the A ADB. 

3. In the diagram of Theorem 60, if A B: AC =8:5, and if the 
fig. P = 8-9 sq. cm., find the areas of the figs. Q and R. 

4. BY and CZ are medians of the triangle ABC, and YZ is 
joined. PMnd the ratio of the triangle BGC to the triangle YGZ. 
[See p. 98.] 

5. ABC is an isosceles triangle, the equal sides AB, AC each 
measuring 3-6". From a point D in AB, & straight line DE is 
drawn cutting AC produced at E, and making the triangle ADE 
equal in area to the triangle ABC. If AD = 1-8", find AE. 

6. AB is a, diameter of a circle, and two chords .IP, AQ aro' 
produced to meet the tangent at B in A' and Y. 

Shew that (i) the A A PQ, A YX are similar ; 

(ii) the four points P, Q, Y, X are concyclic. 

7. In the triangle ABC, the angle A is externally bisected by a 
line wjiich meets the base produced at D and the circum-circlo at 
E ; show that 

AB-AC = AEAD. 

8. Draw an isosceles triangle equal in area to a triangle ABC, 
and having its vertical angle equal to the angle .1. 

9. On a given base draw an isosceles triangle equal in area to a 
given triangle A BC. 

10. Any regular polygon inscribed in a circle is the geometric 
mean between th(! inscribc'd and circumscribed regular polygons of 
half the number of sides. 



SIMILAR FIGURES 



251 



Problem 38 

To draw a figure similar to a given rectilineal figure, and 
equal to a given fraction of it in area. 




Let ABODE be the given figure, to which a similar figure 
is to be drawn, having its area a given fraction (saj- three- 
fourths) of that of the fig. ABODE. 

Construction. Make AF three-fourths of AB. Prob. 7. 

From AB cut off AB' the mean proportional between AF 

and AB. Prob. 39. Note. 

On AB' draw the fig. AB'O'D'E' similar to the fig. ABODE. 

Prob. 40. 
Then the fig. AB'O'D'E' = f of the fig. ABODE. 
Proof. By construction, AB"^ = AFAB. 
Now the figs. ABODE, AB'O'D'E' are similar, and AB, 
AB' are corresponding sides ; 

. ^^. AB'O'D'E' ^ AB'^ 
" &g. ABODE ' AB- 

^ AFAB 

AB'- 
^AF^S 
AB 4 



Theor. 59. 



252 GEOMETRY 



EXERCISES 

1. Divide a triangle ABC into two parts of equal area by a line 
XY drawn parallel to the base BC and cutting the other sides at A' 
and 1'. 

Find (i) by ealeulation, (ii) by measurement, the ratio AX : AB. 

2. Divide a triangle .1 BC into three parts of equal area by lines 

PQ, XY drawn paralh^l to the base BC. If P and X lie in AB, 

prove that 

AP ^AX ^ AB_ 

1 V2 Vf 

Hence shew how a triangle may be divided into n equal parts by 
lines drawn parallel to one side. 

3. Draw a rectangle of length 8 em., and breadth o cm. Then 
draw a similar rectangle of one-third the area. 

Measure its length to the nearest millimetre, and verify your re- 
sult by calculation. 

4. Draw a quadrilateral A BCD from the following data: 
the Z A = 90°; AB = BC = 8 cm. ; AD = DC = G cm. 
Draw a similar quadrilateral to contain an area of 30 sq. cm., 

and find to the nearest millimetre tht^ length of the side; correspond- 
ing to AB. 

5. Di\'ide a circle of radius 3" into three equal parts by means 
of two concentric circles. 

0. Draw a rectilineal figure equal in area to a given figure E, ami 
similar to a given figure S. [Euclid VI. 25.] 

[First replace the given figures E and S by equivalent squares 
(see Problems 10 and 33). Let the sides of these squares be a and 
h respectively, and let s be one of the sides of 5. 

Find p, a fourth proportional to b, a, s, so that h: a = s: p. 

On p draw a figure P similar to the figure S, so that p and n are 
corresponding sides. Then /' is the figure recpiired ; 

e P V- n'- E 

lor — = -i- = — = — 

.S s" fc2 S 

:. the fig. P - the fig. E.\ 



BISECTOR OF VERTICAL ANGLE 253 

MISCELLANEOUS THEOREMS 

* Theorem 61 

If the vertical angle of a triangle is bisected by a straight line 
which cuts the base, the rectangle contained by the sides of the 
triangle is equal to the rectangle contained by the segments of the 
base, together with the square on the straight line which bisects 
the angle. ^^/^ 




Let ABC be a triangle, having the Z BAC bisected bj' AD. 
It is required to prove that 

the red. AB, AC = the red. BD, DC + the sq. on AD. 
Suppose a circle circumscribed about the A ABC ; and 
let AD be produced to meet the O"® at E. 

Join EC. 
Proof. Then in the A BAD, EAC, 

because the Z BAD = the Z EAC, 
and the Z ABD = the Z AEC in the same segment ; 
.'. the remaining Z BDA = the remaining Z EC A ; 
that is, the A BAD, EAC are equiangular to one another ; 

.-. ^ = ^. _ Theor. 50. 

AE AC 

Hence ABAC = AEAD = {AD + DE) AD 

= AD"" -\- ADDE. 
But ADDE = BDDC ; Theor. 56. 

.-. the rect. AB, AC = the rect. BD, DC + the sq. on AD. 

Q.E.D. 



254 GEOMETRY 

* Theorem 62 

If from the vertical angle of a triangle a straight line is drawn 
perpendicular to the base, the rectangle contained by the sides of 
the triangle is equal to the rectangle contained by the perpendicu- 
lar and the diameter of the circum-circle. 




In the A ABC, let AD be the perp, from A to the base 
BC ; and let AE be a diameter of the circum-circle. 
It is required to prove that 

the red. AB, AC = the red. AE, AD. 
Join EC. 
Proof. Then in the A BAD, EAC, 
the rt. angle BDA = the rt. angle EC A, in the semi-circle 
EC A, 

and the Z ABD = the Z AEC, in the same segment ; 
.'. the remaining Z BAD = the remaining Z EAC ; 
that is, the A BAD, EAC are equiangular to one another. 
.: AB : AE = AD : AC ; Theor. 50. 

Hence the rect. AB, AC = the rect. AE, AD. q.e.d. 

Note. Let a, h, c denote the sides of (lie A ABC, R its oinnmi- 
radius, and p the perp. A D. 

Thensinee AEAD = AB-AC 

2Rp = cb. 
he 



"-■u 



■nltc. _ ahe. _ 
2ap ~ 4a" 



D1AG(JNALS OF QUADKILATEllAL 255 

Theorem 63. [Ptolemy's Theorem] 

The rectangle contained by the diagonals of a quadrilateral 
inscribed in a circle is equal to the swn of the two rectangles con^ 
tained by its ojjposite sides. *^- 



A 




Let A BCD be a quadrilateral inscribed in a circle and let 
AC, BD be its diagonals. 
It is required to prove that 

the rect. AC, BD = the rect. AB, CD + the red. BC, DA. 

Make the Z DAE equal to the Z BAC ; 

to each add the Z EAC, 

then the Z DAC = the Z EAB. 

Proof. Since the Z EAB = the Z DAC, 

and the Z ABE = the Z ACD in the same segment ; 

.*. the A EAB, DAC are equiangular to one another ; 

.-. BA:CA=BE:CD; Theor. 50. 

hence ABCD = AC-BE (i) 

Again in the A DAE, CAB, 
the Z DAE = the Z CAB, 
and the Z ADE = the Z ACB, in the same segment ; 
.'. the A DAE, CAB are equiangular to one another; 
.-. DA :CA=DE : CB; 

hence BCDA = ACDE (ii) 

Adding the equal rectangles on each side in (i) and (ii) 
AB- CD + BC- DA = AC- BE + AC- DE 
= AC (BE + DE) 

= AC-BD. Q.ED. 



256 GEOMETRY 

EXERCISES 

1. ABC is an isosceles triangle, and on the base, or l)ase pro- 
duced, any point A' is taken ; shew that the circumscribed circles of 
the triangles ABX, ACX are equal. 

2. From the extremities B, C of the base of an isosceles triangle 
ABC, straight lines are drawn perpendicular to AB, AC respec- 
tivelj', and intersecting at D ; shew that 

BC-AD = 2ABDB. 

3. If the diagonals of a quadrilateral inscribed in a circle are at 
right angles, the sum of the rectangles contained by the opposite 
sides is double the area of the figure. 

4. A BCD is a quadrilateral inscribed in a cii'cle, and the 
diagonal BD bisects AC; shew that 

ADAB = DC- CB. 

5. If the vertex A of a triangle ABC is joined to any point in 
the. base, it will divide the triangle into two triangles such that their 
circumscribed circles have radii in the ratio of AB to AC. 

6. Construct a triangle, having given the base, the vertical 
angle, and the rectangle contained by the sides. 

7. Two triangles of equal area are inscribed in the same circle ; 
shew that the rectangl(> contained by any two sides of the one is to 
the rectangle contained bj' any two sides of the other as the base of 
the second is to the base of the first. 

8. P ia a point on the arc BC of the circum-circle of an equi- 
lateral triangle ABC. If P is joined to A, B, and C, shew that 

PB + PC = PA. 

9. A BCD is a quadrilateral inscribed in a circle, and BD bi- 
sects the angle ABC; if the points A and C are fixed on the cir- 
cumference of the circle, and B is variable in position, shew that 

AB + BC : BD is a. constant ratio. 

10. From the formula R =— (see Note, p. 254) find the 

4A 

value of R when the sides of the triangle are as follows: 
(i) 21'', 20", 13"; (ii) 30 ft., 25 ft.. 11 ft. 
Draw to a convenient scale and check your work by measurement. 



^ EXERCISES IN REVIEW 257 

MISCELLANEOUS EXAMPLES 

PARTS I-IV 

1. The bisector of the angle P of the triangle PQR meets 
QR at S and QR is produced to T. Prove the sum of the angles 
PQR and PRT equals twice the angle PSR. 

2. L and M are the middle^ points of the sides PQ, PR of the 
A PQR. RL and QM are produced to T and *S' so that RL = LT 
and QM = MS. Prove that T, P, S are collinear and that 
PT = PS. 

3. In the isosceles A PQR, PQ = PR. PS and PT are equal 
parts cut off from PQ, PR respectively. QT, RS intersect at 0. 
Prove ^ TOS, QOR isosceles. 

4. A St. line PR is bisected at Q. From P and R PT, RS 
are drawn perpendicular to any other st. line and QS, QT joined; 
prove A Q TS isosceles. 

5. PQR is a A. PS is ± QR and PT bisects angle QPR. 
Prove angle SPT = half the difference of the angles Q and R. 

6. Find a point such that its distances from two given inter- 
secting straight lines shall be equal to two gi^'en lengths. 

7. G is any point in the base EF of the isosceles A DEF. 
DG is joined and bisected at H: Prove HF > IIG. 

8. The vertical Z .4 of the A ABC is bisected by AD which 
meets the base BC at D. DM, DN drawn || to AB, AC resp. 
meet AC in M and AB in A'. Prove the four sides of figure AN DM 
equal. — 

9. The base BC of the A ABC is produced to D. BO bisect- 
ing Z ABC. and CO bisecting Z ACD meet at 0. Prove Z BOC 

= \ A A. 

10. AD joins the vertex A of the triangle ABC to the middle 
point D of BC. Shew that AD>, = > or < BD according as 

Z BAC\s> acute, right, or obtuse. 

s 



258 GEOMETRY 

11. BC is the base of an isosceles triangle ABC. A r]rr\v with 
centre C and radius CB cuts AB, AC in D and E resp. Shew that 
DE is parallel to the bisector of Z B. 

12. The quadrilateral formed by the bisectors of the angles of 
any quadrilateral is cyclic. 

13. PQ and RS are two equal straight lines not in the same 
straight line. Find a point T so that the A 7'PQ = A TRS. 

14. PQRS is a parallelogram. DE drawn W PR meets SP, 
SR produced if necessary at D and E. Prove AQDP = A 
QER. 

15. Trisect a parallelogram by st. lines through a vertex. 

16. P and Q are tw^o fixed points. Find a point such that OP^ 
+ OQ- may be a minimum. 

17. PQRS is a parallelogram. PT is drawn to any point T 
in QR and is any point in PT. Prove A QOR = A TOS. 

18. If two chords of a circle intersect at right angles, the sum 
of the squares on their segments equals the square on a diameter. 

19. Find a point within a given triangle at which the three 
sides subtend equal angles. When Is the solution possible? 

20. Through an intersection of two given circles draw the 
greatest possible st. line terminated by the two circumferences. 

21. Describe a circle of given radius to touch tw-o given circles. 

22. Describe a circle of given radius to touch two given inter- 
secting St. lines. 

23. From a given point P without a given circle draw a secant 
PQR such that PQ = QR. 

24. From the extremities of the diameter of a circle perpen- 
diculars are drawn to any chord. Sliew that the centre is equally 
distant from the feet of the perpendiculars. 

25. Draw a tangent to a circle which shall bisect a given par- 
allelogram which is- outside the circle. 

2G. Describe a circle with given radius to touch a given st. 
line and have its centre in another given st. line. 

27. Describe a circle with given radius to pass through a given 
point and touch a given st. line. 



EXERCISES IN REVIEW 259 

2S. D{ scribe a circle with given radius to touch a given circle 
and a giva st. line. 

29. AD and AE bisect the interior and exterior angles at A 
of A ^ BC, and meet BC at D and E ; and O is the middle point 
of BC. Prove OC^ = OD- OE. 

30. In a given circle inscribe a triangle whose sides are parallel 
to three given st. lines. 

31. Two circles whose centres are .4 and B touch externally at 
P, and CPD is drawn meeting the circles in C and D. Shew that 
the triangles APD, CPB are equal in area. 

32. Construct a triangle equiangular to a given triangle and 
having a given circle for one of its escribed circles. 

33. Construct a triangle, given the base, the vertical angle, 
and the radius of the inscribed circle. 

34. If two circles intersect and through a point on their common 
chord produced two secants are drawn, one to each circle, the 
four points of section of the secants with the circles are concyclic. 

35. If ABC is a triangle, right-angled at A, and AD is drawn 
perpendicular to BC, shew that 

(i) BC^: BA^ = BC : BD; 
(ii) BC'2: C.42 = BC-.CD. 
Hence deduce BC"^ = BA"^ + .4C-. 

36. A triangle ABC is bisected by a straight hne XY drawn 
parallel to the base BC. Determine the ratio AX: AB. 

Hence bisect a triangle by a line drawn parallel to the base. 

37. If two circles have external contact at A, and a common 
tangent, touching them at B and C, meets the line of centres at S, 

A SB A: A SAC = SB:SC. 

38. Two circles intersect at yl jiid B, and at A tangents are 
drawn, one to each circle, meeting the circumferences at C and D. 
If AB, CB, and BD are joined, shew that 

ACBA: A ABD = CB: BD. 

39. DEF is the pedal triangle of the triangle .4 BC ; prove that 

A .4SC: A DBF = AB'-: DB'-; 
fig. AFDC: A DBF = AD'-: BD'-. 



2G0 GEOMETRY 

40. In a tjivon triangle ABC a second triangle is inscribed by 
joining the middle points of the sides. In this inscribed triangle a 
third is inscribed in like manner, and so on. What fraction is the 
fourth triangle of the triangle ABC? 

41. A semi-cii*cle is described on ^B as diameter, and any two 
chords AC, BD are drawn intersecting at P. Shew that 

AB^ = ACAP + BDBP. 

42. Two circles intersect at B and C, and the two direct com- 
mon tangents AE and DF are drawn; if the common chord is 
produced to meet the tangents at G and //, shew that 

GH^ = AE^ +5C2. 

43. If from an external point P, a secant PCD is drawn to a 
circle and PM is perpendicular to a diameter AB, shew that 

PilP = PCPD + AM MB. 

44. Two circles whose centres are C and D intersect at A and 
B; and a straight line PAQ is drawn through A and terminated 
by the circumferences : prove that 

(i) the Z PBQ = the Z CAD; 
(ii) the Z B PC = the Z BQD. 

45. AB is a given diameter of a circle, and CD is any chord 
parallel to AB ; if X is any point in A B, 

AX'2 + XD^ = XA- + XB\ 

46. If the opposite sides of a cyclic quadrilateral are prcduced 
to meet, the bisectors of the angles so formed are perpendicular. 

47. Given the vertical angle, one of the sides containing it, 
and the length of the perpendicular from the vertex on lh<^ base: 
construct the triangle. 

48. A, B, C are three points in order in a straight line: find a 
point P in the straight line such that PA : PB = PB : PC. 

49. Through D, any point in the base of a triangle ABC. 
straight lines DE, DF are drawn parallel to the sides AB, AC, and 
meeting the sides at E, F: shew that the triangle AEF is a mean 
proportional between the triangles FBD, EDC. 

50. Given the base, and the position of the bisector of the 
vertical angle : construct the triangle. 



ANSWEliS 2G1 



ANSWERS TO NUMERICAL EXERCISES 

Since the utmost care cannot ensure absolute accuracy in graphical work, re- 
sults so obtained are likely to be only approximate. The answers here given are 
those found by calculation, and being true so far as they go, furnish a standard 
by which the student may test the correctness of his drawing and measurement. 
HrKults within one per cent of those given in the Answers may usually be con- 
■fldercd satitsfaclory. 

Exercises. Page 15 

;. :U)°; 12(i°; 201°; 80°. 11 min. ; 37 miu. 

2. 112.1"; i5,jO. 5 hrs. 45 min. S. .)()°; 8 hrs. 40 min. 
4. (i) 14.j°, 35°, 145°. (ii) 55°, 5,"^°. 86°, 94°. 

Exercises. Page 27 

1. 08°, .37°, 75° V. nearly. :.\ 00 cm. .{. 2-2", 50°. 73° nearly. 
•J. 37 ft. 6. 101 metres. 7. 27 ft. S. 424 yds., nearly ; N. W. 
.9. 281 yds., 1.35 yds., 1.53 yds. 10. 214 yds. 

Exercises. Page 41 

/. 125°, 55°, 125°. 12. 15 sees., 30 sees. 

Exercises. Page 43 

3. 21°. /,. 'l7°. .;. 92°, 40°. 0. 07°, 02°. 

Exercises. Page 45 

/. 30°, 00°, 90°. 2. (i) .30°, 12°, 72°; (ii) 20°, 80°. 80°. 

3. 40°. 4. 51°, 111°, 18°. o. (1)34°; (ii) 107°. 

6. 08°. 7. 120°. c9. 36°, 72°, 108°, 144°. 

9. 165°. 11. o, 15. 

Exercises. Page 47 

2. (i) 45°; (ii) 30°. 3. (i) 12; (ii) 15. 



262 



GEUMKTRY 



Exercises. Page 54 

4. (i) Sl°; c. (ii) 55°. 



10. 



Degrees 


15° 


30° 


45° 


60° 


75° 


Cm. 


41 


4-6 


5-7 


8-0 


15-6 



11. 



Degrees 
Cm. 


0° 


30° 


60° 


90° 


120° 


150° 


180° 


10 


2-0 


3- 6 


50 


61 


6-8 


70 



37 ft. 



13. 112 ft. 



14. 346 yds. 693 yds. 



14. 
IS. 



2. 
9. 
11. 

12. 



Exercises. Page 61 

54°, 72°, 54°. 15. 36°. 

(i) 16; (ii) 45°; (iii) 111° per sec. 



IG. 4. 



Exercises. Page 68 

6-80 cm. S. 2-24". 4- 0-39. 5. 2-54. 
3-35". 10. 20 miles; 12-6 km. 

147 miles ; 235 km. 1 em. represents 22 km. 
1" represents 15 mi. ; 1" represents 20 mi. 



8. 10-6 cm. 



3. 0-43 in. 



Exercises. Page 79 

. 1-3 em. 5. 2-4". 



Exercises. Page 84 

/. 4-3 cm., 5-2 cm., 61 cm. 2. 110. 3. 200 yards. 

4. 65°, 77 m., 61 m., 56 m. ,1. 6()4 kiKxts. S, 15° K. nearly. 
(J. Results equal. 9 cm. 7. 4-3 cm. ; 98 cm., 60°; 120°. 

5. (i) One solution; (ii) two; (iii) one, right-angled; (iv) im- 

possible. 
0. 380 yds. to. 6-5 cm. ;/. 6-9cm. 

IJ. Two solutions; 10-4 cm. or 4-5 /6". 2-8 cm., 4-5 cm., 5-3 cm. 

cm. 
IS. 5-8 cm., 4-2 cm. U). 7 cm., 8 cm. 

Exercises. Page 89 

/. 60°, 120°. 2. 3-r>4". 3. 2-12". .{. 4-4 cm. 

6. 10-4 cm., 3-4%. 6. 90°. 7. (i)4-25"; (ii) B = D = 90.^ 



ANSWERS 



2G3 



0. 

1',. 

16. 
20. 
23. 



6 sq. in. 
3- 30 sq. in. 
10,000 sq. m 
900 sq. yds. 
1 cm. = 10 yds 
100 sq. ft. 
288 sq. ft. 



Exercises. 
2. 6 sq. in. 
6. 3-36 sq. in 
10. 110 sq. ft. 
48 yds.; 4-8". 
17. l-CV. 
21. 156 sq. ft. 
2Ji. 72 sq. ft. 



Page 104 

3. 2- SO .sq. in. 
7. 198 sq. in. 
') em. 



15. 
IS. 



4- 
s. 

12. 



3r)0 sq. in. 
42 sq. ft^ 
2-6 in. 



11,700 sq. m. 

000 sq. ft. 19. 1154 sq.ft. 

22. 110 sq. ft. 

25. 75 sq. ft. 



Exercises. Page 107 
(i) 22 cm.; (ii) 3-()". 2. 3-4 sq. in. 3. 574-5 sq. in. 

1-5". 5. 1-93", 75°. 

Exercises. Page 109 

(i) ISOsq. ft. ; (ii) 8-4sq. in. ; 1 heetaro. 

(i) 13-44sq. cm. ; (ii) 15-40sq. em. ; (iii) 20- .50 sq. cm. 

15 sq. cm. 4- 6-3 sq. in. 

(i) 8"; (ii) 13 em. G. 3-30 sq. in. 



11,400 sq. yds. 
2.4 cm. ; 5.1 cm, 



Exercises. Page 112 

2. 0312 sq. m. 
4. 2.04"; 2.20". 



Angle 


0° 


30° 


60° 1 90° 


120° 


150° 


180° 


Area in sq. cm. 





7-5 


130 ' 15-0 


130 


7.S , ] 



1. 00 sq. ft. 
If. 132 sq. em. 



/. sq. in. 
5. 31-2 sq. cm. 



1. (i) 25-5 sq. em. 

2. (i) 8-95 sq. in. ; 

4- 3-3 sq. in. 



Exercises. Page 113 

2. 84 sq. yds. 
5. 180 sq. ft. 

Exercises. Page 115 
2. 170 sq. ft. 3. 015 sq. 
G. 5-20 sq. in. 

Exercises. Page 117 
(ii) 15-6 sq. era. 
(ii) 9-5 sq. in. 



m. 



3. 120 sq. m. 
G. 306 sq. m. 

/,. 8-4 sq. in. 
7. 24 sq. cm. 



Exercises. Page 118 



3. 12,500 sq. m. 



5. 7-5 cm. 



6. 3-0 sq. in. 



264 GEOMETRY 

Exercises. Page 123 

1. (i) T) cm. ; (ii) (i-ocin. ; (iiij :i~". .i. yv) 1-0"; (ii) 2-8 cm. 
S. 41ft. /,. 05 miles. .-7. G-lkm. (I. l(j ft. 
7. 48 m. S. 25 mil(>s. .'^ 73 m. 70. 02 ft. 

Exercises. Page 125 

in. (i) and (iii). //. 2-83". 11. 4-24 cm. ; 18 sq. cm. 

W,. 70-71 sq. m. 1.',. p = 0-93 cm. 

m. (i) 20 cm.; 15 cm.; (ii) 40 cm.; 39 cm. 

17. 35 em. ; 12 cm. ; 300 sq. cm. 

IS. (i) 30 sq. in. ; (ii) 90 sq. ft. ; (iii) 120 sq. cm. ; (iv) 240 sq. yds. 

Ifi. 5-1 cm. nearly. 

Exercises. Page 132 

/. 030 sq. cm. ; 15 cni. 

Exercises. Page 134 

2. 8- 5 cm.; 90°. 3. A circle of radriis cm. 
.',. 5-20". a. 0-25". 

Exercises. Page 136 

1. 71cm. .'i. 40 cm. ■'>. 10". G. 3-1 cm.; 150 sq. cm. 

Exercises. Page 140 

1. 23-90 sq. cm. 2. 8-40 sq. in. 

3. 27-52 .sq. cm. ' /,. 129.8(M) sq. m. 

Exercises. Page 149 

/. 5 cm. 2. 24". 3. 0-0". 0-8". /,. V7 = 20 cm. 

r,. 1 ft. a. O-O sq. in. 7. 0-8". 

Exercises. Page 153 

/. 1-7". 2. 3\/2=42cni. 3. 2v'3 - 3-5 cm. 

J,. 17". 0. 5 cm. 

Exercises. Page 155 

fl. 4 cm. 7. 1-3". 



ANSWERS 265 

Exercises. Page 157 
2. 1-85'. .?. 1()2". 

Exercises. Page 160 
.7. rA". i;. l.(j"; 1.5"; 0.0". 

Exercises. Page 163 
1. 74°, 148', Ki''. .?. 11.')". 230°. 3. fjo", S°, 47°. 



Exercises. Page 172 

1. S-Ociu. ?. n-r/'. 3. 8-7 f-m. /,. 12", 07°. o. 2-5". 

Exercises. Page 174 

3. 3 cm. and 17 cm. 

Exercises. Page 176 

1. 72°, 108°, 108°. 

Exercises. Page 180 

S. 1-6". 3. 1-7". 4- 1-98", 1-0". 

Exercises. Page 193 

2. 2-3 cm., 4-6 cm., 0-9 cm. 3. 1-.39". 

4. 0-9 cm. ; 20-78 sq. cm. 7. 3-2 cm. 

Exercises. Page 194 

1. 2-12"; 4-. 50 sq. in. .;. 8-5 cm. t. 2-0". 

Exercises. Page 195 

/,. 128^; 1-7.3". 

Exercises. Page 186 

1. 3-46"; 400". 2. 2.")9S sq. cm. 

4. (!) 41-.57 sq. cm. ; (ii) 77-25 sq. cm. 



266 GEOMETRY 

Exercises. Page 200 

L (i) 28-3 cm. ; (ii) 628-3 cm. e. (i) 10-62 sq. in. ; (ii) 352-99 sq. in. 
S. 11-31 cm.; 10- 18 sq. cm. /,. 56 sq. cm. S. 43-98 sq. in. 

7. 30-5 sq. cm. S. 8-9". .9. 4" ; 3". 10. 12-57 sq. in. 

Exercises. Page 209 

1. (i) 35; (ii) 8; (iii) a. 

3. 4-0", 5-6". 4. 16-5 em., 120 cm. 

5. 4-0 cm., 2-4 cm. ; 160 cm., 90 em. 

Exercises. Page 214 

1. (i) each =3:2; (ii) each =5:3; (iii) each = 5:2. 

8. (i) 1-4"; (ii) 0-8"; (iii) 0-4 cm., 2-4 cm. 

3. (i) 5-6 cm. ; (ii) 7-7 cm., 2-8 cm. 

Exercises. Page 215 

/. 0-9", 0-0"; 4-5", 30"; 3:2. 

2. 20 cm., 1-5 em. ; 140 em., 10-5 em. 

Exercises. Page 217 

/. 10-5 sq. in. 2. 30 cm. 3. 04 sq. cm. 

4. 110". 5. 33-9 acres. 

Exercises. Page 222 

/. (i) 1-2"; (ii)2-0"; (iii) 7-7 cm. 2. (1)2-1"; (ii) 0-3 cm. 

3. QB = 3-5", BR = 2-5". /,. 3-2 cm., 4-2 cm. 
r,. 2-1", 1-8". 6. 5 ft., 12J ft., ^ ft. 
7. 1-2", 1-3", 1-95." <"?. 5f cm. 

9. 0-8 cm., 1-4 cm., 2-1 cm. 

Exercises. Page 230 
/. \. 2. 20 sq. ft. 3. 10 sq. em. J,. 7:5. H. 5-0". 

Exercises. Page 234 

1. 26". 2. 48 ft.; 8 ft. 3. 2 em. ; 32 oni. 

J,. 3-6". 5. 8100 miles; 10 miles. 



ANSWERS 267 

Exercises. Page 238 

1. (i) 10"; (ii) 0-9"; (iii) (i-O cm. 

,?. 1-4", 0-6"; 3-5", 1-5". 3. (i) 2-0; (ii) 2-8; (iii) 20. 

J,. 1-6 cm., 2-4 cm., 3-2 cm. 5. 1-8", 1-2", 0'9". 6. 2-7". 

7. (i) 1-73; (ii)3-16; (iii) 1-G7. 8. (i) 3; (ii)3-21; (iii) 2-26. 

9. (i) 1-2", 1-6", 20"; (ii) 3-0 cm., 3-6 cm., 4-5 em. ; 

(iii) 2-5 cm., 4-3 cm., .50 cm.; (iv) h = 3-4", c = 2-1", nearly. 

10. 140 m., IGO m. ; 125 m. //. 24 ft., 2 ft. 4 in. 

12. (iOft. 13. 72 ft. U,. lOGft.- 

Exercises. Page 245 

3. 0-52. 5. 31:28, noarly. 

Exercises. Page 248 

3. 1.5-48 sq. in. J,. 3-6 m., 1-.5 m. 

B. 90 acres. C. 512 acres, 

7. 1 cm. represents 15 metres. 

Exercises. Page 250 

3. 2-5 sq. cm., G-4 sq. cm. 4. 4: 1. 

5. 7-2". S. G-2 cm., 3-8 em. 

Exercises. Page 252 
1. 1:V2. 3. 4- Gem. .',. G-9 em. 

Exercises. Page 256 
10. (i) 10|"; (ii) ISfft. 



y^ 



^ 



V 






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lijL 



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^X tJh^ fj^-' 






\) 



I " ^./ ^ -J 



^.yCL. 



x^ 



y-S^ 



.,- "^ 



-u>-' 



^^V V ^>IN.CUVV^ 



CA^LrC 



.^tyyci^ c/ 






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